Graphing
Graphing Parabolas:
x-intercepts
Summary
• A parabola is a graph of a quadratic equation.
• The general equation is: y = ( x + a )( x + b)
e.g. y = ( x + 3)( x + 1)
• a and b are the two x-intercepts. They are the
negative of the numbers in the equation.
e.g. x-intercepts are -3 and -1
• The vertex of the parabola is the turning point and is
half way between the x-intercepts.
e.g. The vertex will lay on the line x = -2
• Substitute the x value of the vertex into the original
equation to find the y value of the vertex.
e.g. y = (−2 + 3)(−2 + 1)
= (1)(−1) = −1
The vertex is at (-2,-1)
• An equation written y = x( x + 2) is the same. The x-intercepts are just 0 and -2 this time.
Note: A negative sign at the front of the equation means the parabola is upside down.
For a complete tutorial on this topic visit www.learncoach.co.nz
Old NCEA Questions
1. For the graph below give:
a. the intercepts
b. the function
3. Sketch the graph of the equation:
y = (4 − x)( x + 2)
2. For the graph below, give:
a. the x and y intercepts
b. the equation of the graph.
4. On the grid below, sketch the graph of
y = −( x − 2)( x + 4)
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Practice Questions
For the following equations factorise (if necessary), write out the x-intercepts and vertex location, and plot the graphs.
There are blank graphs on the next page to draw on.
1.
y = ( x − 1)( x + 3)
2.
y = x( x − 3)
3.
y = x 2 − 10 x + 21
4.
y = 9 − x2
5.
y = ( x + 5)( x + 7)
6.
y = (3 − x)( x − 4)
7.
y = x2 + 6x
8.
y = ( x + 3)(2 − x)
For the following graphs give the x-intercepts and write out the equation.
9.
10.
11.
12.
13.
14.
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Graphing
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Answers
NCEA
Practice
1.
5. x-intercepts at x = 1 and x = -3
a. x-intercepts are at x = 1 and x = -2
b.
y-intercept is at y = 2.
(Achieved)
We know the intercepts so sub into equation
giving: y = ( x − 1)( x + 2) . As the graph
is upside down it is necessary to add a
negative to the front of the equation.
The final equation is therefore:
y = −( x − 1)( x + 2)
= − x 2 − x + 2 Vertex is midway between x-intercepts at x = -1. The
y value is: y = ( −1 − 1)( −1 + 3)
= −4
(Merit)
2. a. x-intercepts are at x = -3 and x = 3
y-intercept is at y = -9
(Achieved)
b. From the intercepts:
y = ( x − 3)( x + 3) or y = x 2 − 9 (Achieved)
3. It is a negative parabola with x-intercepts at x = -2
and x = 4 and a y-intercept at y = 8.
Vertex is midway between x-intercepts at x = 1. The
y value is: y = ( 4 − 1)(1 + 2)
= (3)(3)
=9
6. x-intercepts at x = 0 and x = 3
Vertex is midway between x-intercepts at x = 1.5.
The y value is: y = 1.5(1.5 − 3)
= −2.25
(Achieved)
4. It is a negative parabola with x-intercepts x = 2 and x
= -4 and y-intercept y = 8.
Vertex is midway between x-intercepts at x = -1. The
y value is: y = −( −1 − 2)( −1 + 4)
= −(−3)(3)
=9
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(Achieved)
(Achieved)
2
7. y = x − 10 x + 21
= ( x − 7)( x − 3)
x-intercepts at x = 7 and x = 3
Vertex is midway between x-intercepts at x = 5. The
y value is: y = (5 − 7)(5 − 3)
= −4
(Achieved)
(Achieved)
Graphing
8. y = 9 − x 2
= − ( x 2 − 9)
= −( x − 3)( x + 3)
2
11. y = x + 6 x = x ( x + 6)
x-intercepts at x = 3 and x = -3
Vertex is midway between x-intercepts at x = 0. The
y value is: y = −(0 − 3)(0 + 3)
x-intercepts at x = 0 and x = -6
Vertex is midway between x-intercepts at x = -3. The
y value is: y = −3( −3 + 6)
= −9
=9
(Achieved)
9. x-intercepts at x = -7 and x = -5
Vertex is midway between x-intercepts at x = -6. The
y value is: y = ( −6 + 5)( −6 + 7)
(Achieved)
12. x-intercepts at x = 2 and x = -3
Vertex is midway between x-intercepts at x = -0.5.
The y value is: y = ( −0.5 + 3)( 2 − ( −0.5))
= 6.25
= −1
(Achieved)
10. x-intercepts at x = 3 and x = 4
Vertex is midway between x-intercepts at x = 3.5.
The y value is: y = (3 − 3.5)(3.5 − 4)
= 0.25
(Achieved)
13. x-intercepts at x = -8 and x = -5
Equation is therefore: y = ( x + 5)( x + 8)
(Merit)
14. x-intercepts at x = 2 and x = -4
Equation is therefore: y = ( x − 2)( x + 4)
(Merit)
15. x-intercepts at x = 3 and x = 9
Upside down so negative in front.
Equation is therefore: y = −( x − 3)( x − 9)
(Merit)
16. x-intercepts at x = 0 and x = 3
Equation is therefore: y = x ( x − 3)
(Merit)
17. x-intercepts at x = 0 and x = -4
Upside down so negative in front.
Equation is therefore: y = − x ( x + 4)
(Merit)
(Achieved)
18. x-intercepts at x = -1 and x = 4
Equation is therefore: y = ( x + 1)( x − 4)
(Merit)
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