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Question 672959: a)f(x)=x2-8x+12. b) f(x)=4x-x2
i) find where the parabola cuts the y-axis
ii) find the qaudratic equations roots
iii) find the axis of symmetry. and the turning point
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! a)f(x)=x2-8x+12. b) f(x)=4x-x2
i) find where the parabola cuts the y-axis
ii) find the quadratic equations roots
iii) find the axis of symmetry. and the turning point
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a)f(x)=x2-8x+12
complete the square.
y=(x^2-8x+16)+12-16
y=(x-4)^2-4
This is a parabola that opens upwards with vertex at (4,-4)
..
i) find where the parabola cuts the y-axis
set x=0
y=16-4=12
parabola cuts the y-axis at 12 (y-intercept)
..
ii) find the quadratic equations roots
set y=0
(x-4)^2-4=0
(x-4)^2=4
x-4=±√4=±2
x=4±2
roots are:
x=6 and 2 (x-intercepts)
..
iii) find the axis of symmetry. and the turning point.
axis of symmetry: x=4
turning point: (4,-4) (vertex)
I will let you do b)
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