1. Lake Zurich High School
Graphing Parabolas
Using the Vertex
Axis of Symmetry
& y-Intercept
By: Jeffrey Bivin
Lake Zurich High School
Last Updated: October 15, 2007

2. Graphing Parabolas
With your graphing calculator, graph each of the following quadratic equations and identify the vertex and axis of symmetry.
y = x2 + 4x - 7
y = 2x2 + 10x + 4
y = -3x2 + 5x + 9
Jeff Bivin -- LZHS

3. Graph the following parabola
x = -2
y = x2 + 4x - 7
vertex:
axis of symmetry:
(0, -7)
(-2, -11)
y-intercept:
Jeff Bivin -- LZHS

4. Graph the following parabola
y = 2x2 + 10x + 4
vertex:
axis of symmetry:
y-intercept:
Jeff Bivin -- LZHS

5. Graph the following parabola
y = -3x2 + 5x + 9
vertex:
axis of symmetry:
y-intercept:
Jeff Bivin -- LZHS

6. Graphing Parabolas
Now look at the coefficients of the equation and the value of the axis of symmetry – especially a and b
y = ax2 + bx + c
y = x2 + 4x - 7
y = 2x2 + 10x + 4
y = -3x2 + 5x + 9
Jeff Bivin -- LZHS

7. Graphing Parabolas y = ax2 + bx + c Axis of symmetry: Vertex:
Jeff Bivin -- LZHS

8. Graph the following parabola
x = -2
re-visited
y = x2 + 4x - 7
axis of symmetry:
vertex:
(0, -7)
(-2, -11)
y-intercept:
Jeff Bivin -- LZHS

9. Graph the following parabola
re-visited
y = 2x2 + 10x + 4
(0, 4)
axis of symmetry:
vertex:
y-intercept:
Jeff Bivin -- LZHS

10. Graph the following parabola
Why did this parabola open downward instead of upward as did the previous two?
re-visited
y = -3x2 + 5x + 9
axis of symmetry:
vertex:
y-intercept:
Jeff Bivin -- LZHS

11. Graph the following parabola
y = x2 + 6x - 8
x = -3
Axis of symmetry:
(0, -8)
Vertex:
(-3, -17)
y-intercept:
Jeff Bivin -- LZHS

12. Graph the following parabola
y = -2x2 + 7x + 12
(0, 12)
Axis of symmetry:
Vertex:
y-intercept:
Jeff Bivin -- LZHS

13. Graphing Parabolas
In Vertex Form
Jeff Bivin -- LZHS

14. Graphing Parabolas
With your graphing calculator, graph each of the following quadratic equations and identify the vertex and axis of symmetry.
vertex
axis of sym.
y = x2
y = (x - 5)2 - 4
y = -3(x + 2)2 + 5
y = ⅜(x - 3)2 + 1
Jeff Bivin -- LZHS

15. Graph the following parabola
x = 5
y = (x - 5)2 - 4
(0, 21)
axis of symmetry:
vertex:
(5, 4)
y-intercept:
Jeff Bivin -- LZHS

16. Graph the following parabola
(-2, 5)
y = -3(x + 2)2 + 5
axis of symmetry:
vertex:
(0, -7)
y-intercept:
x = -2
Jeff Bivin -- LZHS

17. Graph the following parabola
x = 3
y = ⅜•(x - 3)2 - 1
axis of symmetry:
vertex:
y-intercept:
(3, -1)
Jeff Bivin -- LZHS

18. Graphing Parabolas
In Intercept Form
Jeff Bivin -- LZHS

19. Graph the following parabola
x = 1
y = (x – 4)(x + 2)
x-intercepts:
(-2, 0)
(4, 0)
(0, -8)
axis of symmetry:
(1, -9)
vertex:
y-intercept:
Jeff Bivin -- LZHS

20. Graph the following parabola
x = 5
y = (x - 1)(x - 9)
(0, 9)
x-intercepts:
(1, 0)
(9, 0)
axis of symmetry:
(5, -16)
vertex:
y-intercept:
Jeff Bivin -- LZHS

21. Graph the following parabola
y = -2(x + 1)(x - 5)
(2, 18)
(0, 10)
x-intercepts:
(-1, 0)
(5, 0)
axis of symmetry:
x = 2
vertex:
y-intercept:
Jeff Bivin -- LZHS

22. Convert to standard form
y = -2(x + 1)(x - 5)
y = -2(x2 – 5x + 1x – 5)
y = -2(x2 – 4x – 5)
y = -2x2 + 8x + 10
Jeff Bivin -- LZHS

23. Now graph from standard form.
y = -2x2 + 8x + 10
(2, 18)
Axis of symmetry:
(0, 10)
Vertex:
x = 2
y-intercept:
Jeff Bivin -- LZHS

24. VERTEX 27.50 137.50 Income = Price ● Quantity
A taxi service operates between two airports transporting 200 passengers a day. The charge is $ The owner estimates that 10 passengers will be lost for each $2 increase in the fare. What charge would be most profitable for the service? What is the maximum income?
VERTEX
Income = Price ● Quantity
Define the variable
x = number of $ price increases
f(x) = ( x ) ( 200 – 10x )
15 + 2x = 0
200 – 10x = 0
f(x) = income
2x = -15
200 = 10x
Vertex is:
So, price = (15 + 2x) = (15 + 2(6.25)) = = $27.50
Maximum income:
27.50
137.50
Jeff Bivin -- LZHS

25. Alternative Method VERTEX Income = Price ● Quantity
A taxi service operates between two airports transporting 200 passengers a day. The charge is $ The owner estimates that 10 passengers will be lost for each $2 increase in the fare. What charge would be most profitable for the service? What is the maximum income?
VERTEX
Income = Price ● Quantity
Define the variable
x = number of $ price increases
f(x) = ( x ) ( 200 – 10x )
f(x) = 3000 – 150x + 400x – 20x2
f(x) = income
f(x) = – 20x x
f(6.25) = – 20(6.25) (6.25)
f(6.25) =
Vertex is:
So, price = (15 + 2x) = (15 + 2(6.25)) = = $27.50
Maximum income = f(x) = $
Jeff Bivin -- LZHS