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Question 1 of 4
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The value of aa is positive, so the parabola is concave up. Solve for the yy-intercept by substituting x=0x=0 into the equation.
yy |
== |
3x2+6x3x2+6x |
|
== |
3(02)+6(0)3(02)+6(0) |
Substitute x=0x=0 |
yy |
== |
00 |
Mark the yy-intercept on the graph.
Next, solve for the xx-intercepts by substituting y=0y=0.
yy |
== |
3x2+6x3x2+6x |
00 |
== |
3x2+6x3x2+6x |
Substitute y=0y=0 |
00 |
== |
3x(x+2)3x(x+2) |
Factor out 33 |
3x(x+2)3x(x+2) |
== |
00 |
3x3x |
== |
00 |
Equate factors to 00 |
xx |
== |
00 |
x+2x+2 |
== |
00 |
Equate factors to 00 |
xx |
== |
-2−2 |
Mark the xx-intercepts on the graph.
Draw a parabola using the points.
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Question 2 of 4
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Rewrite the equation so that it is in standard form.
yy |
== |
8x-2x28x−2x2 |
yy |
== |
-2x2+8x−2x2+8x |
The value of aa is negative, so the graph is concave down.
Find the yy-intercept of the equation by substituting x=0x=0.
yy |
== |
−2x2+8x−2x2+8x |
|
== |
−2(02)+8(0)−2(02)+8(0) |
Substitute x=0x=0 |
yy |
== |
00 |
Next, solve for the xx-intercepts by substituting y=0y=0.
yy |
== |
−2x2+8x−2x2+8x |
00 |
== |
−2x2+8x−2x2+8x |
Substitute y=0y=0 |
00 |
== |
-2x(x-4)−2x(x−4) |
Factor out -2x−2x |
-2x(x-4)−2x(x−4) |
== |
00 |
-2x−2x |
== |
00 |
Equate factors to 00 |
xx |
== |
00 |
x-4x−4 |
== |
00 |
Equate factors to 00 |
xx |
== |
44 |
Mark the xx-intercepts on the graph.
Find the vertex from the formula x=-b2ax=−b2a
xx |
== |
-b2a−b2a |
|
|
== |
-82(-2)−82(−2) |
a=-2a=−2,b=8b=8 |
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|
== |
-8-4−8−4 |
|
xx |
== |
22 |
Solve for the yy-intercept of the vertex using the obtained value of xx.
yy |
== |
-2x2+8x−2x2+8x |
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== |
-2(22)+8(2)−2(22)+8(2) |
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== |
-2(4)+16−2(4)+16 |
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== |
-8+16−8+16 |
yy |
== |
88 |
Draw a parabola using the points, together with the obtained vertex (2,8)(2,8)
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Question 3 of 4
Graph y=2x2-4x+3y=2x2−4x+3.
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The value of aa is positive, so the graph is concave up.
Find the vertex of the parabola by first solving for xx from the formula x=-b2ax=−b2a.
xx |
== |
-b2a−b2a |
|
|
== |
--42(2)−−42(2) |
a=2a=2,b=-4b=−4 |
|
|
== |
4444 |
|
xx |
== |
11 |
Substitute the value of xx into the quadratic equation.
yy |
== |
2x2−4x+32x2−4x+3 |
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== |
2(12)−4(1)+32(12)−4(1)+3 |
Substitute x=1x=1 |
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== |
2(1)-4+32(1)−4+3 |
|
== |
2-12−1 |
yy |
== |
11 |
Simplify |
This corresponds to a vertex of (1,1)(1,1). Mark this on the graph.
Find the yy-intercept of the graph. This is equal to cc, which is equal to 33. Plot this on the graph as well.
Connect the points and take note that the parabola opens up.
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Question 4 of 4
Graph y=-x2+4x+5y=−x2+4x+5.
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The value of aa is negative, so the graph is concave down.
Find the yy-intercept of the graph. This is equal to cc, which is equal to 55. Plot this on the graph.
Find the xx-intercept of the graph by substituting y=0y=0.
yy |
== |
-x2+4x+5−x2+4x+5 |
00 |
== |
-x2+4x+5−x2+4x+5 |
y=0y=0 |
Since the equation is in standard form ((aax2+x2+bbx+x+cc=0)=0) we can factorise using the cross method.
To factorise, we need to find two numbers that add to 44 and multiply to 55
Read across to get the factors.
Solve for the xx-intercepts and plot them on the graph.
-x+5−x+5 |
== |
00 |
Solve for xx |
xx |
== |
55 |
x+1x+1 |
== |
00 |
Solve for xx |
xx |
== |
-1−1 |
Connect the points and take note that the parabola opens up.