Why does the vertex of y = a(x-h)^2 + k have h with the opposite sign from what's in the parentheses?

The form is written with a subtraction built in. The squared term is zero exactly when x - h = 0, that is, when x = h. So if you see (x+6)^2, ask what makes the inside zero: x = -6. That's why the vertex is at -6 even though you see a plus sign. Rewriting (x+6)^2 as (x-(-6))^2 makes the match to the pattern obvious.

How do I know whether the vertex is a maximum or a minimum without graphing?

Look only at the sign of a. If a is positive the parabola opens upward like a cup, the arms head up forever, and the vertex is the lowest point — a minimum. If a is negative it opens downward and the vertex is the highest point — a maximum. The size of a and the location of the vertex don't affect this at all.

Is c in y = ax^2 + bx + c the same as k in vertex form?

No. The constant c is the y-intercept, the height of the graph at x = 0. The value k is the height at the vertex, x = h. They're equal only when the vertex is on the y-axis, which happens when b = 0. In y = x^2 - 6x + 4 the y-intercept is 4 but the vertex is at (3,-5).

How many points do I need to plot a good parabola?

Five is plenty: the vertex, two points on one side, and their two mirror images. Always include the vertex, since it fixes the curve's position, and always use symmetry rather than substituting into the equation twice as many times. Connect them with a smooth curve that flattens at the vertex — never with straight line segments.

ALG1-8.1

Quadratic Functions: Graphs & Vertex Form

Learn to graph quadratics: find the vertex and axis of symmetry from y = a(x − h)² + k or x = −b/(2a), tell which way the parabola opens, and read max/min in context.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Quadratic Functions: Graphs & Vertex Form, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Quadratic Functions: Graphs & Vertex Form free

What this lesson covers

Every quadratic function graphs as a parabola — a smooth U-shaped curve that is perfectly symmetric about a vertical line. Once you know three things (which way it opens, where its turning point sits, and how wide or narrow it is), you can sketch the whole graph from just a few points. That turning point, the vertex, is the star of this lesson. It's the highest or lowest point on the curve, so in real situations it answers questions like "what is the maximum height?" or "what number of items gives the lowest cost?"

In this lesson you'll read the vertex straight off vertex form

y = a(x-h)^2 + k

, compute it from standard form

y = ax^2 + bx + c

using

x = -\frac{b}{2a}

, and use symmetry to finish a graph fast. These skills come back in every remaining lesson of the unit.

Parabolas, Direction of Opening, and Width

A quadratic function is any function you can write as

y = ax^2 + bx + c

with

a \neq 0

. The

x^2

term is what bends the graph; without it you'd have a line.

The leading coefficient

a

controls two things at once. Its sign tells you the direction of opening, and its absolute value tells you how narrow the parabola is compared to the parent function

y = x^2

Value of $a$	Opens	Vertex is a	Shape vs. $y=x^2$
$a > 0$	Upward	Minimum	—
$a < 0$	Downward	Maximum	Reflected over the $x$ -axis
$\lvert a \rvert > 1$	—	—	Narrower (stretched vertically)
$0 < \lvert a \rvert < 1$	—	—	Wider (compressed vertically)

Every parabola has an axis of symmetry: the vertical line through the vertex. Points on the graph come in mirror pairs at equal horizontal distances from that line, and they share the same

y

-value. That symmetry is a huge time-saver — plot one side, reflect it, done.

A very common misconception is that

a

moves the graph up or down. It does not. Changing

a

from

1

3

keeps the vertex in place and squeezes the arms inward. The graph moves only when

h

k

change. Another frequent slip: assuming a bigger

\lvert a \rvert

makes a wider curve. It's the opposite — a larger

\lvert a \rvert

means

y

grows faster for the same

x

, so the parabola climbs more steeply and looks narrower.

Reading the Vertex from Vertex Form

Vertex form is

y = a(x-h)^2 + k

, and the vertex is the point

(h, k)

. The axis of symmetry is

x = h

.

The form is built so that the squared quantity equals zero exactly when

x = h

. Since

a(x-h)^2

is never negative when

a>0

(and never positive when

a<0

), the smallest (or largest) possible value of

y

happens at that one input, and there

y = k

. That is the whole reason

(h,k)

is the turning point.

The number one error here is the sign of

h

. The form has a minus sign built into it, so you must match the pattern exactly.

Equation	Rewrite to match $a(x-h)^2+k$	Vertex
$y = (x-5)^2 + 2$	already matches	$(5, 2)$
$y = (x+5)^2 + 2$	$y = (x-(-5))^2 + 2$	$(-5, 2)$
$y = -4(x+1)^2 - 7$	$y = -4(x-(-1))^2 + (-7)$	$(-1, -7)$
$y = 3x^2 + 6$	$y = 3(x-0)^2 + 6$	$(0, 6)$

Notice that

k

keeps its sign but

h

flips. A good habit: say out loud "

x

minus what makes this zero?" For

(x+5)^2

, you need

x = -5

, so

h = -5

.

Once you have the vertex, get one more point by plugging in any convenient

x

, then mirror it across the axis. For

y = 2(x-3)^2 - 1

, the vertex is

(3,-1)

; at

x = 4

y = 2(1)-1 = 1

, so

(4,1)

is on the graph and so is its mirror

(2,1)

Finding the Vertex from Standard Form

When a quadratic is given as

y = ax^2 + bx + c

, the vertex is not visible, but symmetry hands it to you. The axis of symmetry is

x = -\frac{b}{2a}

This is the

x

-coordinate of the vertex. To get the

y

-coordinate, substitute that value back into the original equation. Do not try to read

k

off the equation —

c

is the

y

-intercept, not the vertex height. Those two are equal only when

b = 0

.

Work carefully with signs. In

y = x^2 - 6x + 4

b = -6

(the sign travels with the coefficient), so

x = -\frac{-6}{2(1)} = 3

. Then

y = 9 - 18 + 4 = -5

, giving vertex

(3,-5)

.

The formula isn't magic. The

x

-intercepts of a parabola, when they exist, are symmetric about the axis, and averaging the two roots from the quadratic formula gives exactly

-\frac{b}{2a}

. The axis sits halfway between any two points with the same

y

-value.

That fact gives a useful shortcut for a third and fourth point: the

y

-intercept is always

(0, c)

, and its mirror image is

\left(-\frac{b}{a},\, c\right)

— the same distance on the other side of the axis. For

y = x^2 - 6x + 4

, the

y

-intercept

(0,4)

mirrors to

(6,4)

, since the axis

x=3

is three units from each.

Where students most often go wrong: forgetting the negative sign in front of the fraction, or dividing by

a

instead of

2a

. Both mistakes put the axis in the wrong place and wreck every point after it.

Interpreting the Vertex as a Maximum or Minimum

In an applied problem, the vertex answers "how high," "how low," "how much," or "when." Keeping the two coordinates straight is the key skill.

The

x

-coordinate answers when or at what input the extreme occurs. The

y

-coordinate answers what the extreme value is. If

h(t) = -16t^2 + 48t + 6

gives a ball's height in feet after

t

seconds, then

t = -\frac{48}{2(-16)} = 1.5

seconds is when the ball peaks, and

h(1.5) = -36 + 72 + 6 = 42

feet is how high it gets. Answering "the maximum height is 1.5" is the single most common mistake on problems like this.

Because

a = -16

is negative, the parabola opens downward and the vertex is a maximum. If

a

were positive — say a cost function

C(x) = 2x^2 - 40x + 500

— the vertex would be a minimum, and

x = 10

would be the production level that makes cost as small as possible.

Also think about domain in context. A ball's height only makes sense for

t \geq 0

and until it lands, and you can't produce a negative number of items. The full parabola extends forever in both directions, but the meaningful part of the graph is usually a piece of one or both arms plus the vertex.

One more reading skill: the range. If the vertex is a minimum at

(h,k)

, the range is

y \geq k

. If it's a maximum, the range is

y \leq k

. The domain of an unrestricted quadratic is all real numbers either way.

Putting a Full Graph Together

A reliable five-step routine produces an accurate sketch every time.

First, check the sign of

a

so you know which way the curve opens. Second, find the vertex — read

(h,k)

from vertex form, or compute

x=-\frac{b}{2a}

and substitute from standard form. Third, draw the dashed axis of symmetry through the vertex. Fourth, find two or three extra points on one side of the axis, using the

y

-intercept

(0,c)

if you're in standard form. Fifth, reflect those points across the axis and connect everything with a smooth curve.

Step	Vertex form $y=a(x-h)^2+k$	Standard form $y=ax^2+bx+c$
Direction	sign of $a$	sign of $a$
Vertex	$(h,k)$ read directly	$x=-\frac{b}{2a}$ , then substitute
Axis	$x=h$	$x=-\frac{b}{2a}$
Easy point	plug in $x=h+1$	$y$ -intercept $(0,c)$
Mirror point	reflect over $x=h$	reflect over the axis

Two habits keep sketches honest. Connect points with a smooth curve, not straight segments — a parabola never has a corner, and the flattest part is right at the vertex. And make the two arms symmetric; if your left arm rises faster than your right, you've either misplaced the vertex or plotted a point incorrectly.

Finally, a parabola may cross the

x

-axis twice, touch it once, or miss it entirely. That depends on where the vertex sits relative to the

x

-axis and which way the curve opens: a vertex above the axis on a curve opening upward never crosses. You'll quantify this later in the unit with the discriminant, but you can already see it in a graph.

Key terms

Parabola.: The U-shaped graph of a quadratic function, symmetric about a vertical line through its turning point.
Vertex.: The turning point of a parabola, written $(h,k)$ ; the lowest point when the graph opens upward and the highest point when it opens downward.
Axis of symmetry.: The vertical line $x = h$ (equivalently $x = -\frac{b}{2a}$ ) that divides the parabola into two mirror-image halves.
Vertex form.: $y = a(x-h)^2 + k$ , the form in which the vertex $(h,k)$ can be read directly from the equation.
Standard form.: $y = ax^2 + bx + c$ with $a \neq 0$ ; here $c$ is the $y$ -intercept and the vertex must be computed.
Leading coefficient.: The value $a$ ; its sign gives the direction of opening and its absolute value controls how narrow or wide the parabola is.
Maximum value.: The largest output of a function; for a quadratic with $a<0$ it equals the $k$ -value of the vertex.
Minimum value.: The smallest output of a function; for a quadratic with $a>0$ it equals the $k$ -value of the vertex.

Worked example

For

y = 2x^2 - 8x + 5

, find the direction of opening, the vertex, the axis of symmetry, and the

y

-intercept. State whether the vertex is a maximum or a minimum, give the range, and list one mirror pair of points to help sketch the graph.

Step 1 — Direction. Here

a = 2

b = -8

c = 5

. Since

a > 0

, the parabola opens upward, so the vertex will be a minimum.

Step 2 — Axis of symmetry. Use

x = -\frac{b}{2a} = -\frac{-8}{2(2)} = \frac{8}{4} = 2

. The axis of symmetry is the line

x = 2

. Watch the double negative:

b

-8

, not

8

.

Step 3 — Vertex. Substitute

x = 2

into the original equation:

y = 2(2)^2 - 8(2) + 5 = 8 - 16 + 5 = -3

. The vertex is

(2, -3)

. Notice that

-3

is not

c

; the

y

-value of the vertex almost never equals the constant term.

Step 4 — Minimum and range. Because the graph opens upward, the minimum value of the function is

-3

, occurring at

x = 2

. The range is

y \geq -3

, and the domain is all real numbers.

Step 5 — Extra points using symmetry. The

y

-intercept is

(0, 5)

. It sits 2 units left of the axis

x=2

, so its mirror is 2 units right:

(4, 5)

. For one more pair, try

x = 1

y = 2 - 8 + 5 = -1

, giving

(1,-1)

and its mirror

(3,-1)

.

Step 6 — Sketch and check. Plot

(2,-3)

(1,-1)

(3,-1)

(0,5)

(4,5)

and join them with a smooth curve. As a check, rewrite in vertex form:

y = 2(x-2)^2 - 3

. Expanding gives

2(x^2-4x+4)-3 = 2x^2-8x+8-3 = 2x^2-8x+5

, which matches the original.

Practice questions

What is the vertex of

y = -3(x+4)^2 + 7

, and which way does the graph open?

Vertex $(4, 7)$ , opens upward
Vertex $(-4, 7)$ , opens downward
Vertex $(-4, -7)$ , opens downward
Vertex $(4, -7)$ , opens upward

Answer: Vertex $(-4, 7)$ , opens downward

Match the equation to

y = a(x-h)^2 + k

. Since

x+4 = x-(-4)

, we get

h = -4

, so the vertex sits at

x=-4

, not

x=4

— the

h

-value flips sign. The constant

+7

k

and keeps its sign, so the vertex is

(-4, 7)

. Because

a = -3

is negative, the parabola opens downward and

7

is the maximum value of the function.

A quadratic function has the axis of symmetry

x = 5

and passes through the point

(2, 11)

. Give another point that must be on the graph, and explain your reasoning.

Answer: $(8, 11)$

Points on a parabola with the same

y

-value are equally distant from the axis of symmetry. The point

(2,11)

5 - 2 = 3

units to the left of the line

x=5

, so its mirror image is 3 units to the right, at

x = 5 + 3 = 8

. Reflecting across a vertical line changes only the

x

-coordinate, so the

y

-value stays

11

, giving

(8, 11)

. This reflection trick lets you double your plotted points without any extra substitution.

A firework is launched so that its height in feet after

t

seconds is

h(t) = -16t^2 + 96t + 10

. When does it reach its greatest height, and what is that height?

Answer: It peaks at $t = 3$ seconds at a height of 154 feet.

Because

a = -16 < 0

, the parabola opens downward and the vertex is a maximum. The time of the peak is the

x

-coordinate:

t = -\frac{b}{2a} = -\frac{96}{2(-16)} = -\frac{96}{-32} = 3

seconds. Substitute to get the height:

h(3) = -16(9) + 96(3) + 10 = -144 + 288 + 10 = 154

feet. The most common error is reporting 3 as the maximum height; 3 is the time, and the height is the

y

-coordinate. Note also that

10

is the launch height at

t=0

, not the peak.

FAQ

Why does the vertex of $y = a(x-h)^2 + k$ have $h$ with the opposite sign from what's in the parentheses?: The form is written with a subtraction built in. The squared term is zero exactly when $x - h = 0$ , that is, when $x = h$ . So if you see $(x+6)^2$ , ask what makes the inside zero: $x = -6$ . That's why the vertex is at $-6$ even though you see a plus sign. Rewriting $(x+6)^2$ as $(x-(-6))^2$ makes the match to the pattern obvious.
How do I know whether the vertex is a maximum or a minimum without graphing?: Look only at the sign of $a$ . If $a$ is positive the parabola opens upward like a cup, the arms head up forever, and the vertex is the lowest point — a minimum. If $a$ is negative it opens downward and the vertex is the highest point — a maximum. The size of $a$ and the location of the vertex don't affect this at all.
Is $c$ in $y = ax^2 + bx + c$ the same as $k$ in vertex form?: No. The constant $c$ is the $y$ -intercept, the height of the graph at $x = 0$ . The value $k$ is the height at the vertex, $x = h$ . They're equal only when the vertex is on the $y$ -axis, which happens when $b = 0$ . In $y = x^2 - 6x + 4$ the $y$ -intercept is $4$ but the vertex is at $(3,-5)$ .
How many points do I need to plot a good parabola?: Five is plenty: the vertex, two points on one side, and their two mirror images. Always include the vertex, since it fixes the curve's position, and always use symmetry rather than substituting into the equation twice as many times. Connect them with a smooth curve that flattens at the vertex — never with straight line segments.

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