1. The standard form of a quadratic equation is y = ax2 + bx + c.

2. The graph of a quadratic equation is a parabola.

3. When a > 0, the graph opens up.

4. When a < 0, the graph opens down.

5. Every parabola has a vertex. For graphs opening up, the vertex is a minimum (low point). For graphs opening down, the vertex is a maximum (high point). The y-value is the minimum or maximum value!

6. The x-coordinate of the vertex is equal to .

To find the x coordinate of the vertex, use the equation

Then substitute the value of x back into the equation of the parabola and solve for y.

xb

a

2

You are given the equation y=-x2 + 4x –1. Find the coordinates of the vertex.

xb

a= -

2

x = 2

-= =a b1 4,

( -x = - 4

2 1)( )

2y x x= - + -4 1

y = - + -4 8 1

y =3

y = - + –22 4 2 1( ) ( )

The coordinates of the vertex are (2,3)Substitute and solve for y

Standard Form FeaturesLet f(x) = ax2 + bx + c, where a ≠ 0.

• If |a| > 1, the parabola becomes narrower.

• If |a| < 1, the parabola becomes wider.• The axis of symmetry is x = -b/2a and passes through

the graph’s vertex.• The vertex has the coordinates (-b/2a, f(-b/2a)).

• (0, c) is the y-intercept for the graph.

Choose two values of x that are to the right or left of the x-coordinate of the vertex.

Substitute those values in the equation and solve for y.

Graph the points. (Keep in mind the value of a as this will help you determine which way the graph opens.)

Since a parabola is symmetric about the vertical line through the vertex, you can plot mirror image points with the same y-values on the “other side” of the parabola.

x y = -x2 + 4x – 1 y

1

-1

y = -(1)2 + 4(1) – 1

y = -1 +4 – 12

y = -(-1)2 + 4(-1) – 1

y = -1 – 4 – 1-6

You can check these solutions by comparing the y value with the y-value for the symmetric x-values (1 3, -15) (3,2), (5,-6)

x

yPlot the vertex and the points from your table of values: (2,3), (1,2), (-1,-6).

Use the symmetry of parabolas to plot two more points on the “other side” of the graph. The point (1,2) is one unit away from the line of symmetry, so we can also plot the point (3,2). The point (-1,-6) is three units away from the line of symmetry, so we can also plot the point (5,-6).

Sketch in the parabola.

Korean terms / US termsStandard Form / Vertex Form 기본형 y = a(x-h)2 + k (h,k) is the vertex

General Form / Standard Form 일반형 y=ax2 +bx + c (0,c) is the y-intercept

Forms

ExampleGiven the equation f(x) = -1/2(x - 1)2 + 3,

determine:Opens up or down?

Vertex: ________

Axis of Sym.: _____

Sketch the graph.a = - not 1½

a < 0 ( - ) means opens down1 means over 1 up 1-1 means over 1 down 1 over 2 down 4 - means over 1 down ½ ½

over 2 down 2

y

x

5

5

-5

-5

(1,3)

x=1

Standard form to Vertex form

For the function f(x) = x2 - 2x - 3, find:1. X-intercepts: _______

4. Y-intercept: ________

5. The equation

in vertex form:

____________________

6. Opens up or down?___

3. Vertex: _______

2. Axis of Sym.: ______

y

x

5

5

-5

-5

(-1,0), (3,0) by solving x2 - 2x - 3 =0 and from this we can determine the Axis of Symmetry

(0,-3) the constant in standard form (0,c)

y=(x-1)2 – 4 a=1 in original equation

Up a>0

(1,-4)

x=1 substituting 1

for x into the equation will give us the y-value of the vertex

Using the VertexA soccer goalie kicks the ball

straight up with an initial velocity of 56 feet per second. The equation h = -16t2 + 56t describes the height, h, of the ball t seconds after being kicked.How long until the ball reached its

peak?What is the maximum height?How long was the ball in the air?

h = -16t2 + 56t  

Practice worksheets