How do you find the vertex and the intercepts for f(x)= x^2+10x-8f(x)=x2+10x−8?

1 Answer
Binayaka C.
May 9, 2016

Vertex is at (-5, -33)(−5,−33) x-intercepts are at (0.745,0) & (-10.745,0)(0.745,0)&(−10.745,0)
y intercept is at (0,-8)(0,−8)

Explanation:

f (x) = x^2+10x-8 = (x+ 5 )^2 -33f(x)=x2+10x−8=(x+5)2−33 So vertex is at (-5, -33)(−5,−33) Y intercept:
putting x=0 ; f(x)= -8 :. y intercept is at (0,-8) putting f(x)=0 ; x^2+10x-8=0 :. x=-10/2 +-(sqrt (100+32))/2 :. x=0.745 , -10.745 So x-intercepts are at (0.745,0) & (-10.745,0) graph{x^2+10x-8 [-80, 80, -40, 40]}[Ans]

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