y = -x^2+2x+8
x=0, y =8 (0,8)
when y = 0
x^2-2x-8=0
x^2-4x+2x-8=0
x(x-4)+2(x-4)=0
(x-4)(x+2)=0
x= 4 OR -2
x Intercept points are (4,0) & (-2.0)
so you know what the intercepts are
You can put this solution on YOUR website! f(x)=-x^2+2x+8
Find the x and y intercepts
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replace f(x) with y:
y = -x^2+2x+8
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To find y-intercepts set x=0 solve for y:
y = -x^2+2x+8
y = -(0)^2+2(0)+8
y = 0+0+8
y = 8
y-intercept is at (0,8)
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To find x-intercepts set y=0 solve for x:
y = -x^2+2x+8
0 = -x^2+2x+8
0 = x^2-2x-8
0 = (x-4)(x+2)
x = {4, -2}
x-intercepts are at (-2,0) and (4,0)