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

1 Answer
Binayaka C.
May 9, 2016

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

Explanation:

#f (x) = x^2+10x-8 = (x+ 5 )^2 -33# So vertex is at #(-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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