How do you find the vertex and intercepts for #y=x^2+8x-7#?

1 Answer
Mar 6, 2016

Vertex is at #(-4,-23)# Y intercept is at #(0,-7)# X-intercepts are at #(0.796,0)# and #(-8.796,0)#

Explanation:

#y=x^2+8x-7=x^2+8x+16-23 = (x+4)^2-23# So vertex is at #(-4,-23)#
y-intercept : putting x=0 in the equation we get #y= -7#
x-intercept : Putting y=0 we get #x^2=8x-7 = 0# solving the quadratic equation we get #x=0.796# and #x=-8.796# graph{x^2+8x-7 [-80, 80, -40, 40]}[Answer]