How do I find the axis of symmetry x=__, the x intercepts(_,_),(_,_), and the y-intercept(_,_) for f(x)=x^2+8x-9

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ishpiro | College Teacher | (Level 1) Educator

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`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

 

 

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