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Question 1 of 4
Graph y=−x2+6x−9.
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The value of a is negative, so the graph is concave down.
Solve for the y-intercept by substituting x=0.
y |
= |
−x2+6x−9 |
|
= |
−(02)−6(0)−9 |
Substitute x=0 |
|
= |
0−0−9 |
y |
= |
−9 |
Simplify |
Find the vertex of the parabola by solving for x from the formula x=−b2a.
x |
= |
−b2a |
|
|
= |
−62(−1) |
a=−1,b=6 |
|
|
= |
−6−2 |
|
x |
= |
3 |
This also marks the axis of symmetry.
Substitute the value of x back into the quadratic equation to find the vertex.
y |
= |
−x2+6x−9 |
|
= |
−(32)+6(3)−9 |
Substitute x=3 |
|
= |
−9+18−9 |
y |
= |
0 |
Simplify |
This corresponds to a vertex of (3,0). Mark this on the graph.
Using the vertex and the axis of symmetry obtained above, a graph can now be drawn.
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Question 2 of 4
Graph the function
y=2x2−5x−3
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First, find the x values by factoring using cross method
(2x+1)(x−3) |
= |
0 |
2x+1 |
= |
0 |
2x+1 −1 |
= |
0 −1 |
2x |
= |
−1 |
2x÷2 |
= |
−1÷2 |
x |
= |
−12 |
x−3 |
= |
0 |
x−3 +3 |
= |
0 +3 |
x |
= |
3 |
Mark these 2 points on the x axis
Next, find the axis of symmetry and add it to the graph
x |
= |
−b2a |
Axis of Symmetry |
|
x |
= |
−(−5)2(2) |
Substitute values |
|
x |
= |
54 |
Now, substitute x=54 to the equation to know where the graph intersects the axis of symmetry
y |
= |
2x2−5x−3 |
|
|
= |
2(54)2−5(54)−3 |
Substitute x=54 |
|
|
= |
−498 |
|
y |
= |
−618 |
Finally, substitute x=0 to find where graph intersects the y axis
y |
= |
2x2−5x−3 |
|
= |
2(0)2−5(0)−3 |
Substitute x=0 |
|
= |
0−0−3 |
y |
= |
−3 |
Simply connect the points to form a parabola
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Question 3 of 4
Graph the function
y=−x2+2x+15
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First, find the x values by factoring using cross method
(x+3)(−x+5) |
= |
0 |
x+3 |
= |
0 |
x+3−3 |
= |
0−3 |
x |
= |
−3 |
−x+5 |
= |
0 |
−x+5+x |
= |
0+x |
5 |
= |
x |
x |
= |
5 |
Mark these 2 points on the x axis
Next, find the axis of symmetry and add it to the graph
x |
= |
−b2a |
Axis of Symmetry |
|
x |
= |
−22(−1) |
Substitute values |
|
x |
= |
−2−2 |
|
x |
= |
1 |
Now, substitute x=1 to the equation to know where the graph intersects the axis of symmetry
y |
= |
−x2+2x+15 |
|
= |
−(1)2+2(1)+15 |
Substitute x=1 |
|
|
= |
−1+2+15 |
y |
= |
16 |
Finally, substitute x=0 to find where graph intersects the y axis
y |
= |
−x2+2x+15 |
|
= |
−02+2(0)+15 |
Substitute x=0 |
|
= |
−0+0+15 |
y |
= |
15 |
Simply connect the points to form a parabola
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Question 4 of 4
Graph the function
y=−x2+3x−5
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First, find the x values using the quadratic formula
x |
= |
−b±√b2−4ac2a |
Quadratic Formula |
|
|
= |
−3±√32−4(−1)(−5)2(−1) |
Plug in the values of a,b and c |
|
|
= |
−3±√9−20−2 |
|
|
= |
−3±√−11−2 |
Remember that a negative value inside a surd gives out imaginary roots, hence x has no solution
This means the graph does not touch the x axis
Next, find the axis of symmetry
x |
= |
−b2a |
Axis of Symmetry |
|
x |
= |
−32(−1) |
Substitute values |
|
x |
= |
−3−2 |
|
x |
= |
32 |
Now, substitute x=32 to the equation to know where the graph intersects the axis of symmetry
y |
= |
−x2+3x−5 |
|
|
= |
−94+92−5 |
Substitute x=32 |
|
y |
= |
−234 |
Finally, substitute x=0 to find where graph intersects the y axis
y |
= |
−x2+3x−5 |
|
= |
−02+3(0)−5 |
Substitute x=0 |
|
= |
−0+0−5 |
y |
= |
−5 |
Simply connect the points to form a parabola