Vertex form and the vertex it names directly, graphing from it, modelling a suspension cable, intercept form and the x-intercepts it names, the axis of symmetry halfway between them, and converting between the three forms.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 4 — Quadratic Functions and Factoring
Graph Quadratic Functions in Vertex or Intercept Form
Objectives
Five outcomes. The last one explains why three forms exist at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 245-249 — the lesson these objectives are drawn from
Warm-up
Lesson 4.1 found the vertex of a standard-form quadratic with a formula. This lesson finds it by reading.
Discussion prompt
In Lesson 4.1 you found that y equals 2x squared minus 8x plus 6 has vertex (2, -2). What would you have to write instead of the equation for that vertex to be readable without any calculation?
Hint: Look at Lesson 2.7's absolute value form and copy its shape.
Answer:
\[ y = 2(x - 2)^2 - 2 \]
Expanding confirms it: 2 times x squared minus 4x plus 4, minus 2, is 2x squared minus 8x plus 8 minus 2, which is the original. The 2 and the negative 2 in the brackets ARE the vertex, exactly as h and k were for the V in Lesson 2.7.
Concept
The same parabola can be written in standard form, vertex form or intercept form. They describe one curve, but each puts a different feature in plain sight — the y-intercept, the vertex, or the x-intercepts.
vertex form — The form y equals a times the quantity x minus h, squared, plus k. Its vertex is the point with coordinates h and k.
\[ y = a(x - h)^2 + k \]
Choosing a form is choosing what you want to be able to read off without work. Nothing is gained or lost about the curve itself.
Figure (svg): The same parabola written three ways, with what each form reveals at a glance
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 245-246
Section
Section 1
Concept
In y equals a times the quantity x minus h squared, plus k, the vertex is at the point with coordinates h and k, and the axis of symmetry is the vertical line through it. No formula is needed.
\[ y = a(x - h)^2 + k, \quad \text{vertex } (h, k) \]
This is exactly Lesson 2.7's form with the absolute value replaced by a square, and h behaves the same way: the form subtracts it, so x plus 2 means h equals negative 2.
Figure (svg): A parabola in vertex form with its vertex, axis of symmetry and two reflected points marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 245-245 — Graph of Vertex Form
Picture it
Example 1: y equals negative one quarter times the quantity x plus 2 squared, plus 5.
Figure (svg): A parabola in vertex form with its vertex, axis of symmetry and two reflected points marked
The coefficient gives the width and direction; h and k give the position. Three numbers, three independent facts, no calculation.
Worked example
Example 1, step one. Reading is the whole of it.
\[ \text{Identify } a, h, k \text{ and the vertex of } y = -\tfrac{1}{4}(x + 2)^2 + 5. \]
Read a
Why: The coefficient outside the bracket is negative one quarter. Its size is under one, so the parabola is wider than the parent; its sign is negative, so it opens down.
\[ a = -\frac{1}{4} \]
Read h by asking what makes the bracket zero
Why: x plus 2 is zero at x equal to negative 2, so h is negative 2.
\[ h = -2 \]
Read k
Why: The constant outside the square is 5.
\[ k = 5 \]
State the vertex and axis
Why: The vertex is (-2, 5) and the axis of symmetry is the vertical line x equals negative 2.
\[ \text{vertex } (-2, 5) \]
Figure (svg): The solution to Worked example read the vertex form shown as a ladder of expressions, one row per algebraic move
\[ \text{vertex } (-2, 5), \quad \text{axis } x = -2 \]
Verify: substitute the vertex's x value
Why: At x equal to negative 2 the bracket is zero, so the squared term contributes nothing and y is exactly 5 — the vertex height. And because a is negative, 5 is the largest value the function ever reaches, which confirms the vertex is a maximum.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 245-245
Matching
Ask what makes the bracket zero.
Match the pairs
Why: The first coordinate is always the value making the bracket vanish and the second is always the constant outside. Two of these share the same h and differ only in k, which is a good illustration that the two numbers are read independently and neither affects the other.
Worked example
Guided Practice 1 to 3, read rather than computed.
\[ \text{Find the vertex of } y = (x+2)^2 - 3, \; y = -(x-1)^2 + 5, \; f(x) = \tfrac{1}{2}(x-3)^2 - 4. \]
First function
Why: The bracket is zero at x equal to negative 2 and the constant is negative 3.
\[ \text{vertex } (-2, -3) \]
Second function
Why: Zero at x equal to 1, constant 5, and a is negative so it opens down.
\[ \text{vertex } (1, 5),\text{ opens down} \]
Third function
Why: Zero at x equal to 3, constant negative 4, and a is one half so it is wider than the parent.
\[ \text{vertex } (3, -4),\text{ wider} \]
Note the pattern in the signs
Why: A plus inside the bracket gives a negative h; a minus inside gives a positive h. The constant outside is read as written.
Figure (svg): The solution to Worked example three more vertex forms shown as a ladder of expressions, one row per algebraic move
\[ (-2, -3); \quad (1, 5); \quad (3, -4) \]
Verify: substitute each vertex x back in
Why: In each case the bracket becomes zero, leaving exactly the constant: negative 3, then 5, then negative 4. That is what makes the vertex the extreme value — the squared term can only add to it, or with a negative a only subtract from it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 246-246
Trap
\[ y = -\tfrac{1}{4}(x + 2)^2 + 5 \]
Take h as positive 2 because the bracket shows a plus
Why: The sign inside the bracket is copied straight into the vertex.
\[ \text{vertex } (2, 5) \quad \text{(wrong)} \]
Substituting x equal to 2 gives negative one quarter of 16, plus 5, which is 1 — not the vertex value of 5.
\[ y = -\tfrac{1}{4}(x - (-2))^2 + 5 \]
Find the x that makes the bracket zero
Why: That value is h, and asking the question sidesteps the sign rule entirely.
\[ x + 2 = 0 \;\Longrightarrow\; x = -2 \;\Longrightarrow\; \text{vertex } (-2, 5) \]
Only h flips. The constant outside the square is read exactly as it appears, which is why (-2, 5) has a negative first coordinate and a positive second.
Sorting
Read the coefficient in front of the bracket.
Sort into buckets
Sort each vertex-form function.
Sign for direction, size for width. The same two independent readings as in Lesson 4.1 and Lesson 2.7.
Fill the middle
Guided Practice 3.
Fill in the blanks
f(x) = \tfrac3, -4___(x - 3)^2 - 4 \;\Longrightarrow\; \text___ (___)
Why: The bracket vanishes at x equal to 3, so h is 3, and the constant outside is negative 4, so k is negative 4. The coefficient of one half makes the parabola wider than the parent but does not move the vertex — a affects shape only.
Prediction
Commit before reasoning.
Predict first
In y equals a times the quantity x minus h squared, plus k, which letter behaves differently from the absolute value form of Lesson 2.7?
Correct: None — all three behave identically.
\[ y = a\lvert x - h \rvert + k \qquad \text{versus} \qquad y = a(x-h)^2 + k \]
Why: The coefficient a still gives width and direction, h still shifts horizontally with its sign flipped, and k still shifts vertically as written. The only difference between the two families is the SHAPE the form produces: a V there, a smooth parabola here. Recognising that the three parameters transfer unchanged is what makes each new function family faster to learn than the last.
Section
Section 2
Concept
With the vertex read off directly, graphing needs only a couple of extra points. Evaluate at two convenient x values, plot them, and reflect both across the axis of symmetry.
\[ y = -\tfrac{1}{4}(x + 2)^2 + 5 \]
Choose x values that make the bracket a whole number, and ideally an even one when a is a fraction, so the arithmetic stays clean.
Figure (svg): A parabola in vertex form with its vertex, axis of symmetry and two reflected points marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 245-245 — Graph a quadratic function in vertex form
Picture it
Example 1: the vertex, two evaluated points, and their reflections.
Figure (svg): A parabola in vertex form with its vertex, axis of symmetry and two reflected points marked
The symmetry doubles every point you compute. Two evaluations and one reading give five points, which is more than enough for a confident curve.
Worked example
Example 1, all four steps.
\[ \text{Graph } y = -\tfrac{1}{4}(x + 2)^2 + 5. \]
Read a, h and k and note the direction
Why: a is negative one quarter, so the parabola is wider than the parent and opens down; the vertex is (-2, 5).
Plot the vertex and draw the axis
Why: The axis of symmetry is the vertical line x equals negative 2.
\[ \text{axis } x = -2 \]
Evaluate at two convenient values
Why: At x equal to 0 the bracket is 2, squared is 4, a quarter of that is 1, so y is 4. At x equal to 2 the bracket is 4, squared is 16, a quarter is 4, so y is 1.
\[ (0, 4)\text{ and } (2, 1) \]
Reflect both across the axis
Why: The point (0,4) is 2 right of the axis, so its mirror is 2 left, at (-4,4). Similarly (2,1) mirrors to (-6,1).
\[ (-4, 4)\text{ and } (-6, 1) \]
Figure (svg): The solution to Worked example graph from vertex form shown as a ladder of expressions, one row per algebraic move
\[ y = -\tfrac{1}{4}(x + 2)^2 + 5 \]
Verify: substitute a reflected point
Why: At x equal to negative 4 the bracket is negative 2, whose square is 4, so y is negative 1 plus 5, which is 4 — matching the reflection. The mirror points really do lie on the curve, so the symmetry saved two genuine calculations.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 245-245
Fill the middle
Example 1 at x equal to 2.
Fill in the blanks
y = -\tfrac1___(2 + 2)^2 + 5 = -\tfrac______(16) + 5 = ___
Why: Two plus two is four, squared is sixteen, a quarter of sixteen is four, and the negative sign makes it negative four; adding five gives one. Note the order: the bracket is evaluated first, then squared, then multiplied by a, then the constant is added — which is the order of operations from Lesson 1.2 applied to a formula.
Worked example
Guided Practice 1, graphed the same way.
\[ \text{Graph } y = (x + 2)^2 - 3. \]
Read the three constants
Why: a is 1, so the parent width and opening up; the vertex is (-2, -3).
\[ \text{vertex } (-2, -3),\text{ opens up} \]
Evaluate at two nearby values
Why: At x equal to negative 1 the bracket is 1, so y is 1 minus 3, which is negative 2. At x equal to 0 the bracket is 2, so y is 4 minus 3, which is 1.
\[ (-1, -2)\text{ and } (0, 1) \]
Reflect both
Why: One unit right of the axis mirrors to one unit left, giving (-3,-2); two units right mirrors to (-4,1).
\[ (-3, -2)\text{ and } (-4, 1) \]
Join with a smooth curve
Why: Five points, opening upward from the vertex.
Figure (svg): The solution to Worked example an upward parabola from vertex form shown as a ladder of expressions, one row per algebraic move
\[ y = (x + 2)^2 - 3, \quad \text{vertex } (-2, -3) \]
Verify: confirm the vertex is the minimum
Why: The squared bracket is at least zero for every x, so y is at least negative 3 always, with equality only at x equal to negative 2. That makes negative 3 the minimum value and (-2,-3) the lowest point, exactly as a positive a requires.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 246-246
Error analysis
A student graphs Example 1 and mirrors a point incorrectly.
Annotate
On: \( \text{the axis is } x = -2 \text{ and } (0, 4) \text{ is on the curve, so its mirror is } (-2, 4) \)
Count the horizontal distance from the point to the axis, then count the same distance the other way. Reflecting is about distance, not about landing on the axis.
Ranking
Graphing from vertex form.
Put in order
Why: The vertex must be plotted before the axis can be drawn through it, and the axis must exist before anything can be reflected across it. Evaluating two points rather than one means the curve's width is pinned down as well as its position — one point plus the vertex would leave the shape ambiguous.
Prediction
Commit before reasoning.
Predict first
For y equals negative one quarter times the quantity x plus 2, squared, plus 5, which x values keep the arithmetic cleanest?
Correct: x equal to 0 and 2, giving brackets of 2 and 4.
\[ x = 0: \; -\tfrac{1}{4}(4) + 5 = 4 \qquad x = -1: \; -\tfrac{1}{4}(1) + 5 = 4.75 \]
Why: The coefficient is a quarter, so a bracket whose square is divisible by four keeps the output whole. Brackets of 2 and 4 give squares of 4 and 16, both divisible by four. Brackets of 1 and 3 give 1 and 9, producing quarters and fractional outputs. Any values do work, but choosing multiples that match the denominator of a saves fractions throughout.
Comparison
Fill the blanks. One saves a calculation.
Comparison matrix
| Step | From standard form | From vertex form |
|---|---|---|
| Find the vertex | compute -b/(2a), then substitute | read h and k directly |
| Find the axis | through the computed vertex | x = h |
| Get a free point | the y-intercept c, plus its reflection | evaluate two points and reflect |
| Direction and width | from a | from a |
Vertex form removes the vertex calculation entirely, which is its whole advantage. Standard form gives the y-intercept free, which vertex form does not.
Section
Section 3
Concept
When a real situation has a lowest or highest point — a cable's sag, a ball's peak, a cost's minimum — vertex form puts that point in the equation, so the model can be read rather than solved.
\[ y = \tfrac{1}{7000}(x - 1400)^2 + 27 \]
The symmetry is often as useful as the vertex itself: knowing that two equal heights sit equally far from the axis answers questions the picture does not.
Figure (svg): A suspension bridge cable modelled by a parabola, with the vertex marked and the symmetry giving the tower spacing
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 246-246 — Use a quadratic model in vertex form
Picture it
Example 2: the Tacoma Narrows Bridge, with towers 307 feet above the roadway.
Figure (svg): A suspension bridge cable modelled by a parabola, with the vertex marked and the symmetry giving the tower spacing
The vertex says the cable's lowest point is 1400 feet from the left tower. Symmetry says it must be 1400 from the right one too, so the towers are 2800 feet apart.
Worked example
Example 2. The answer comes from the vertex and the symmetry, not from solving anything.
\[ \text{A cable is modelled by } y = \tfrac{1}{7000}(x - 1400)^2 + 27, \text{ with equal towers. Find their spacing.} \]
Read the vertex
Why: The bracket vanishes at x equal to 1400 and the constant is 27, so the vertex is (1400, 27).
\[ \text{vertex } (1400, 27) \]
Interpret the vertex
Why: The lowest point of the cable is 27 feet above the roadway, 1400 feet from the left tower.
\[ \text{lowest point at } x = 1400 \]
Apply the symmetry
Why: The two towers are the same height, so they are the same distance from the axis of symmetry.
Compute the spacing
Why: Fourteen hundred feet on each side of the vertex.
\[ d = 2800\text{ feet} \]
Figure (svg): The solution to Worked example the distance between the towers shown as a ladder of expressions, one row per algebraic move
\[ d = 2(1400) = 2800 \text{ feet} \]
Verify: check the tower height the model predicts
Why: At x equal to 0 the model gives one seven-thousandth of 1400 squared, plus 27, which is 1,960,000 over 7000 plus 27, or 280 plus 27 — exactly 307 feet, the stated tower height. The model is consistent with the picture, which confirms the reading of the vertex.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 246-246
Real world
A ball's height in feet t seconds after being thrown is h equals negative 16 times the quantity t minus 2, squared, plus 68.
Discussion prompt
State the vertex and say what each coordinate means. When does the ball reach its highest point, how high is it, and what does the negative coefficient tell you?
Hint: One coordinate is a time and the other a height.
Answer:
The vertex is (2, 68): the ball reaches its highest point 2 seconds after being thrown, at a height of 68 feet.
The negative coefficient means the parabola opens downward, so 68 is a maximum — which is exactly what a thrown ball's height should be. A positive coefficient would have described something falling to a minimum and then rising again, which nothing thrown does.
The size, 16, is not arbitrary either: it comes from gravity, and it appears in every such model measured in feet and seconds.
Worked example
Guided Practice 4. Only the coefficient changes.
\[ \text{Compare } y = \tfrac{1}{6500}(x - 1400)^2 + 27 \text{ with the model in Example 2.} \]
Compare the vertices
Why: Both brackets vanish at 1400 and both constants are 27, so both cables have the same lowest point.
Compare the coefficients
Why: One six-thousand-five-hundredth is LARGER than one seven-thousandth, since the denominator is smaller.
\[ \frac{1}{6500} > \frac{1}{7000} \]
Say what a larger coefficient does
Why: A larger size of a makes the parabola narrower, so the cable rises more steeply from its lowest point.
Check what that means physically
Why: At the same horizontal distance the new cable is higher, so with 307-foot towers the towers would be closer together.
Figure (svg): The solution to Worked example compare two cable models shown as a ladder of expressions, one row per algebraic move
\[ \text{same vertex } (1400, 27), \text{ but narrower} \]
Verify: compute both at the same x
Why: At x equal to 0 the original gives 307 feet and the new one gives 1,960,000 over 6500 plus 27, which is about 301.5 plus 27, or 328.5 feet. The new cable is higher at the same distance, confirming it is the steeper of the two.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 246-246
Error analysis
A student answers Example 2 with the vertex's x-coordinate.
Annotate
On: \( \text{the vertex is at } x = 1400, \text{ so the towers are } 1400 \text{ feet apart} \)
In a symmetric model, ask whether the question wants a distance from the axis or across the whole figure. They differ by a factor of two, which is easy to miss and impossible to hide.
Prediction
Commit before reasoning.
Predict first
Two cables have the same vertex but coefficients of one seven-thousandth and one six-thousand-five-hundredth. Which rises more steeply?
Correct: The one with one six-thousand-five-hundredth.
\[ \tfrac{1}{6500} \approx 0.000154 > \tfrac{1}{7000} \approx 0.000143 \]
Why: A smaller denominator makes a larger fraction, and a larger size of a makes the parabola narrower — rising faster away from the vertex. The instinct that 6500 is smaller than 7000 and so must give a shallower curve is the trap: what matters is the size of the coefficient, which is the reciprocal of the denominator.
Socratic
One question, and nothing else on this slide.
\[ y = \tfrac{1}{7000}(x - 1400)^2 + 27 \]
Discussion prompt
This parabola is defined for every real x, but the cable is not. What is the model's real domain, what does it predict outside that range, and how would you know where to stop?
Hint: Ask where the towers are.
Answer:
The cable runs between the two towers, so the model's real domain is x from 0 to 2800 feet. Outside that there is no cable.
At x equal to 4000 the model predicts a height of about 1,041 feet, which is more than three times the tower height — arithmetic without meaning, exactly like the paramotor below ground in Lesson 1.5.
You would know where to stop from the situation, not from the algebra. The equation gives no hint that anything special happens at 0 or 2800; only knowing what it models does.
Fill the middle
Example 2 at the left tower.
Fill in the blanks
y(0) = \tfrac307___(0 - 1400)^2 + 27 = \tfrac______ + 27 = ___
Why: Fourteen hundred squared is 1,960,000, and dividing by 7000 gives 280; adding 27 gives 307 feet — the stated tower height. Checking a model against a fact you were given, rather than one you derived, is what confirms the model was read correctly before you use it for anything else.
Section
Section 4
Concept
In y equals a times the quantity x minus p, times the quantity x minus q, the parabola crosses the horizontal axis at p and at q. The axis of symmetry sits exactly halfway between them.
intercept form — The form y equals a times the quantity x minus p times the quantity x minus q, in which p and q are the x-intercepts of the parabola.
\[ y = a(x - p)(x - q), \quad x = \frac{p + q}{2} \]
Only a parabola that actually crosses the horizontal axis can be written this way. One that stays entirely above or below it has no intercept form at all.
Figure (svg): A parabola in intercept form with both x-intercepts marked and the vertex halfway between them
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 246-247 — Graph of Intercept Form
Picture it
Example 3: y equals 2 times the quantity x plus 3, times the quantity x minus 1.
Figure (svg): A parabola in intercept form with both x-intercepts marked and the vertex halfway between them
The intercepts are negative 3 and 1 — the values making each bracket zero — and the axis is at their average, negative 1.
Worked example
Example 3, all three steps.
\[ \text{Graph } y = 2(x + 3)(x - 1). \]
Find the x-intercepts
Why: Each bracket vanishes at a different value: x plus 3 at negative 3, and x minus 1 at 1.
\[ (-3, 0)\text{ and } (1, 0) \]
Find the axis of symmetry
Why: Halfway between the intercepts: negative 3 plus 1, over 2.
\[ x = -1 \]
Find the vertex height
Why: Substituting x equal to negative 1 gives 2 times 2 times negative 2.
\[ y = -8 \]
Draw the parabola
Why: Through the two intercepts and the vertex at (-1, -8), opening up since a is positive.
\[ \text{vertex } (-1, -8) \]
Figure (svg): The solution to Worked example graph from intercept form shown as a ladder of expressions, one row per algebraic move
\[ \text{vertex } (-1, -8), \quad \text{intercepts } -3 \text{ and } 1 \]
Verify: check both intercepts satisfy the equation
Why: At x equal to negative 3 the first bracket is zero, so the whole product is zero — the curve is on the axis. Same at x equal to 1 for the second bracket. A product is zero exactly when one of its factors is, which is why the intercepts are readable straight from the brackets.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 247-247
Matching
Set each bracket to zero.
Match the pairs
Why: In every case the intercept is the value making that bracket zero, which flips the sign shown. The last one has both brackets showing plus signs and therefore two negative intercepts — a parabola sitting entirely to the left of the vertical axis, which is worth being able to predict from the equation alone.
Worked example
The same procedure with a negative leading coefficient.
\[ \text{Graph } y = -(x + 1)(x - 5). \]
Find the intercepts
Why: The brackets vanish at negative 1 and at 5.
\[ (-1, 0)\text{ and } (5, 0) \]
Find the axis of symmetry
Why: Negative 1 plus 5, over 2.
\[ x = 2 \]
Find the vertex height
Why: Substituting x equal to 2 gives negative one times 3 times negative 3, which is 9.
\[ y = 9 \]
Note the direction
Why: a is negative, so the parabola opens down and the vertex is a maximum.
\[ \text{vertex } (2, 9), a\text{ maximum} \]
Figure (svg): The solution to Worked example a downward parabola in intercept form shown as a ladder of expressions, one row per algebraic move
\[ \text{vertex } (2, 9), \quad \text{intercepts } -1 \text{ and } 5 \]
Verify: check the vertex is above the intercepts
Why: The intercepts are both at height zero and the vertex is at height 9, above them — which is what a downward-opening parabola requires, since the vertex must be the highest point. Had the vertex come out below zero with a negative a, something would be wrong.
Trap
\[ y = 2(x + 3)(x - 1) \]
Read the intercepts as 3 and negative 1, copying the signs in the brackets
Why: The numbers are lifted from the equation without asking what makes each bracket zero.
\[ \text{intercepts } 3 \text{ and } -1 \quad \text{(wrong)} \]
Substituting x equal to 3 gives 2 times 6 times 2, which is 24 — nowhere near zero.
\[ y = 2(x + 3)(x - 1) \]
Set each bracket to zero and solve
Why: The intercepts are the values that make a factor vanish, and a plus inside gives a negative intercept.
\[ x + 3 = 0 \;\Rightarrow\; x = -3 \qquad x - 1 = 0 \;\Rightarrow\; x = 1 \]
The same sign flip as h in vertex form, and the same cure: ask what makes the bracket zero rather than reading the sign.
Fill the middle
Example 3's intercepts.
Fill in the blanks
x = \frac-1___ = \frac______ = ___
Why: Negative three plus one is negative two, and halving gives negative one. The axis of symmetry is always the average of the two intercepts, because the parabola crosses the axis at equal distances either side of its vertex — the same symmetry that reflected points in every previous section.
Prediction
Commit before reasoning.
Predict first
Which of these could NOT be written in intercept form?
Correct: A parabola that never meets the horizontal axis.
\[ y = x^2 + 1 \quad \text{is never zero, so it cannot be factored over the reals} \]
Chapter 4 will return to this: whether a quadratic has real x-intercepts is exactly what the discriminant of Lesson 4.8 measures.
Why: Intercept form is a product of factors, and a product is zero exactly when a factor is zero — so any parabola written that way must reach zero somewhere. A parabola sitting entirely above the axis, such as y equals x squared plus 1, never does, so it has no intercept form. The touching case does have one, with p equal to q, giving a repeated factor.
Explain it to yourself
The formula for the axis in intercept form is an average.
\[ x = \frac{p + q}{2} \]
Discussion prompt
Explain why the axis of symmetry must lie exactly halfway between the two x-intercepts. What property of the parabola forces it?
Hint: The two intercepts have the same height.
Answer:
Both intercepts sit at height zero, so they are two points at the SAME height on the curve. The parabola's symmetry means any two points at equal height are equally far from the axis, one on each side.
Being equally far from the axis on opposite sides is exactly what being halfway between means, so the axis is at their midpoint — which is the average of the two values.
Section
Section 5
Concept
Any vertex or intercept form can be multiplied out into standard form. Going the other way is harder: every quadratic has a vertex form, but only those with real x-intercepts have an intercept form.
\[ 2(x+3)(x-1) = 2x^2 + 4x - 6 \]
Converting is worth doing when the form you have hides what you need. Standard form for the y-intercept, vertex form for the vertex, intercept form for the crossings.
Figure (svg): Two columns showing that expanding always works but factoring back is not always possible
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 245-247
Picture it
Expanding never fails; factoring sometimes does.
Figure (svg): Two columns showing that expanding always works but factoring back is not always possible
Lesson 4.7 will show how to convert standard form into vertex form for any quadratic at all, by completing the square. Intercept form remains conditional.
Worked example
Multiplying out, using the distributive property twice.
\[ \text{Write } y = 2(x + 3)(x - 1) \text{ in standard form.} \]
Multiply the two brackets first
Why: Each term of the first times each term of the second: x times x, x times negative 1, 3 times x, 3 times negative 1.
\[ x ^{2} - x + 3 x - 3 \]
Combine like terms inside
Why: Negative x plus 3x is 2x.
\[ x ^{2} + 2 x - 3 \]
Multiply by the leading coefficient
Why: Two times each term.
\[ 2 x ^{2} + 4 x - 6 \]
Read what standard form now shows
Why: The y-intercept is negative 6, which the intercept form did not display.
\[ \text{y-intercept } -6 \]
Figure (svg): The solution to Worked example intercept form to standard form shown as a ladder of expressions, one row per algebraic move
\[ y = 2x^2 + 4x - 6 \]
Verify: check the vertex using Lesson 4.1's formula
Why: Negative b over 2a is negative 4 over 4, which is negative 1, and substituting gives 2 minus 4 minus 6, which is negative 8. That is the vertex (-1, -8) found from intercept form in Section 4 — two forms, one curve, same vertex.
Sorting
Each form makes exactly one feature readable without work.
Sort into buckets
Sort each feature by the form that displays it directly.
The axis of symmetry is readable from intercept form too, as the average of p and q — but that is one calculation rather than none.
Worked example
Squaring the bracket, then distributing.
\[ \text{Write } y = 2(x + 1)^2 - 8 \text{ in standard form.} \]
Square the bracket
Why: x plus 1, squared, is x squared plus 2x plus 1 — not x squared plus 1.
\[ x ^{2} + 2 x + 1 \]
Multiply by the leading coefficient
Why: Two times each of the three terms.
\[ 2 x ^{2} + 4 x + 2 \]
Subtract the constant
Why: Two minus eight is negative six.
\[ 2 x ^{2} + 4 x - 6 \]
Compare with the previous example
Why: The same standard form, so all three of these are the same parabola.
Figure (svg): The solution to Worked example vertex form to standard form shown as a ladder of expressions, one row per algebraic move
\[ y = 2x^2 + 4x - 6 \]
Verify: check the vertex and intercepts agree
Why: Vertex form said the vertex is (-1,-8); intercept form said the crossings are at -3 and 1; standard form says the y-intercept is -6. Substituting x equal to 0 into any of the three gives -6, and substituting -3 gives 18 minus 12 minus 6, which is 0. All three forms agree on every feature, as they must.
Error analysis
A student converts a vertex form to standard form.
Annotate
On: \( 2(x + 1)^2 - 8 = 2(x^2 + 1) - 8 = 2x^2 - 6 \)
A squared bracket is a product, not a term-by-term operation. Checking one value — here the vertex — catches the missing middle term at once.
Fill the middle
Example 3's function.
Fill in the blanks
2(x + 3)(x - 1) = 2(x^2 + 2x - 3) = 2x^2 + 4x - 6
Why: Two times x squared is 2x squared, two times 2x is 4x, and two times negative 3 is negative 6. The leading coefficient reaches every term inside the brackets — the same distributive discipline as Lesson 1.2, applied to a trinomial rather than a binomial.
Discrimination
Expanding always works; factoring depends on the intercepts.
Sort into buckets
Sort each conversion by whether it is always possible.
Commit first
Answer, then rate your confidence honestly.
Predict first
Do all three forms of the same quadratic have the same value of a?
Correct: Yes — a is the same in all three.
\[ 2(x+3)(x-1) = 2(x+1)^2 - 8 = 2x^2 + 4x - 6 \]
Why: Expanding either factored form leaves the leading coefficient untouched, because the highest-degree term is a times x squared however the rest is arranged. So a means the same thing in every form: the direction of opening and the width. That is a genuinely useful invariant — you can read the shape of a parabola from any of the three forms without converting.
Comparison
Fill the blanks. Each form is a different view of the same curve.
Comparison matrix
| Form | Written as | Reads off directly |
|---|---|---|
| Standard | ax^2 + bx + c | the y-intercept, c |
| Vertex | a(x - h)^2 + k | the vertex, (h, k) |
| Intercept | a(x - p)(x - q) | the x-intercepts, p and q |
| All three | share the same a | the direction and the width |
The last row is what makes converting safe: whatever you do to the arrangement, the leading coefficient survives unchanged.
Pattern
One routine, whichever form arrives.
Step one takes a second and decides the whole approach. Using the standard-form formula on a vertex form is legal and wasteful.
OpenStax Algebra and Trigonometry 2e, §5.1 Quadratic Functions §5.1
Check
Vertex form. Ask what makes the bracket zero.
Check your understanding
What is the vertex of y = -(1/4)(x + 2)^2 + 5?
Answer: A
Why: The bracket vanishes at x = -2, so h is -2, and the constant outside the square is 5, so k is 5. Substituting x = -2 gives y = 5, confirming it.
Check
Intercept form. Set each bracket to zero.
Check your understanding
What are the vertex and x-intercepts of y = 2(x + 3)(x - 1)?
Answer: A
Why: The brackets vanish at -3 and 1. The axis is their average, -1, and substituting gives 2 times 2 times -2, which is -8.
Check
Converting forms. Watch the squared bracket.
Check your understanding
Write y = 2(x + 1)^2 - 8 in standard form.
Answer: A
Why: The bracket squares to x^2 + 2x + 1, multiplying by 2 gives 2x^2 + 4x + 2, and subtracting 8 gives 2x^2 + 4x - 6.
Real world
A firework is launched from the ground, reaches its peak 4 seconds later at 256 feet, and lands back on the ground.
Discussion prompt
Write its height as a function of time in vertex form, then in intercept form, and say which form answered which question more easily. When does it land?
Hint: Launching from the ground gives you one x-intercept immediately.
Answer:
\[ h = a(t - 4)^2 + 256, \quad h(0) = 0 \;\Longrightarrow\; 16a + 256 = 0 \;\Longrightarrow\; a = -16 \]
\[ h = -16(t - 4)^2 + 256 = -16t(t - 8) \]
Vertex form gave the peak height directly — it was written into the equation. Intercept form gives the landing time directly: the factors vanish at t equal to 0 and t equal to 8, so it lands after 8 seconds.
Notice the symmetry doing the work again: it takes as long to fall as to rise, so landing is at twice the peak time. Either form could have answered either question, but each made one of them a matter of reading rather than solving.
Commit first
Answer, then rate your confidence honestly.
Predict first
Can a parabola have exactly one x-intercept?
Correct: Yes — when the vertex sits exactly on the horizontal axis.
\[ y = (x - 3)^2 = (x-3)(x-3) \quad \text{touches at } x = 3 \text{ only} \]
Why: A parabola whose vertex is on the axis touches it at exactly one point rather than crossing. In intercept form that means p equals q, giving a repeated factor such as y equals a times the quantity x minus 3, squared. So there are three possibilities in all — two intercepts, one, or none — which is exactly what the discriminant of Lesson 4.8 will distinguish.
Explain it
They can graph from standard form and wonder why other forms exist.
Discussion prompt
In four sentences or fewer, explain what vertex form and intercept form each give you for free, and why anyone would convert between them.
Hint: Each form saves a particular calculation.
Answer:
Vertex form writes the vertex into the equation, so you read it off instead of computing negative b over 2a. Intercept form writes the x-intercepts into the equation, so you find where the parabola crosses the axis by setting each bracket to zero.
You convert when the form you have hides what you need — expanding to standard form to see the y-intercept, say. All three describe the same curve, so nothing is gained or lost except convenience.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For h and for the intercepts, ask what makes each bracket zero rather than reading the sign. For the vertex from intercept form, average the two intercepts and then substitute. For squaring a bracket, write it as a product of two brackets before multiplying, so the middle term cannot be lost. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Choose one parabola of your own and write it in all three forms, showing the algebra that converts each into standard form. Down the left of the page, list the three forms and beside each write the one feature it reveals without any calculation, circling that feature in your own equations. In the middle, draw the parabola once, marking the vertex, the axis of symmetry, both x-intercepts and the y-intercept, and label each with which form gave it to you. At the bottom, write one quadratic that has NO intercept form and explain in a sentence why not. Finally, in a margin, write the value of a and note that it is the same in all three of your forms.
If your no-intercept-form example does cross the axis, pick one whose vertex is above the axis with a opening upward — such as x squared plus 1 — and check that it never reaches zero.
Recap
Five things, and the last one explains why the first four are worth having.
| If you see | Then |
|---|---|
| A squared bracket | Vertex form: the vertex is (h, k) |
| Two brackets multiplied | Intercept form: the crossings are p and q |
| A plus inside a bracket | The corresponding value is negative |
| A need for the y-intercept | Expand to standard form |
| A parabola never touching the axis | It has no intercept form |
Lesson 4.3 asks the reverse question: given a quadratic in standard form, how do you find the factors that intercept form needs — which is factoring, and it is where the next two lessons go.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.2 Graph Quadratic Functions in Vertex or Intercept Form §4.2, pp. 245-249 — everything on these slides traces back here
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