Question

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

Answer 1

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]