How do I find the axis of symmetry x=__, the x intercepts(_,_),(_,_), and the y-intercept(_,_) for f(x)=x^2+8x-9
`f(x) = x^2 + 8x - 9` is a quadratic function. It's graph is a parabola (see below.)
The axis of symmetry of a parabola is a vertical line passing through its vertex.
For the quadratic function in the form `f(x) = ax^2 + bx+c` the x-coordinate of the vertex is given by `x_0 = -b/(2a)` .
For the given function, a = 1, b =8 and c = -9. So the x-coordinate of the vertex is `x_0 = -8/(2*1) = -4`
So the axis of symmetry is the line with the equation x = 4.
The x-intercepts can be found by solving equation f(x) = 0.
`x^2 + 8x - 9 = 0`
Factor the left side:
(x +9)(x - 1) = 0
x = -9 and x = 1 are the x-intercepts of the graph of this function.
y-intercept is y=f(0) and can be found by plugging in x = 0 into the equation for the function:
`f(0) = 0^2 + 8*0 - 9 = -9`
y-intercept is y = -9