How do you find the vertex and intercepts for #y = x^2 - 8#?
1 Answer
vertex =
y-intercept =
x-intercepts =
Explanation:
The general equation for a quadratic function in vertex form is:
#y = a(x-h)^2+k#
With this equation,
#y=1(x-0)^2-8#
The vertex of a quadratic equation in vertex form is
To find the y-intercept, substitute
#y=x^2-8#
#y=(0)^2-8#
#y=(0)-8#
#y=-8#
To find the x-intercept, substitute
#y=x^2-8#
#0=x^2-8#
#8=x^2#
#x=+-sqrt(8)#
#x=2sqrt(2)# or#-2sqrt(2)#
#x=2.83# or#-2.83#
Here is a graph of the equation:
graph{y= x^2-8 [-12.78, 12.53, -11.14, 2.05]}
As you can see, the graph has a y-intercept of