Answer:
The function is given as [tex]y=(x-1)^{2}-9[/tex]
We will simplify the function,
i.e. [tex]y=x^2-2x+1-9[/tex] gives [tex]y=x^2-2x-8[/tex]
Therefore, the standard form is given by [tex]f(x)=x^2-2x-8[/tex].
As, the given functions have factors (x-4) and (x+2).
Since, the intercept form of a function is its factored form.
Thus, the intercept form is [tex]f(x)=(x+4)(x-2)[/tex].
Now, it is known that,
X-coordinate of the vertex is [tex]\frac{-b}{2a}[/tex] i.e. [tex]\frac{2}{2\times 1}[/tex] i.e. [tex]\frac{2}{2}[/tex] i.e. x= 1
So, the value of y at x=1 is, [tex]y=1^2-2\times 1-8[/tex] i.e. [tex]f(x)=1-2-8[/tex] i.e. y= -9
Thus, the vertex of the given function is (1,-9).
Since, 'the y-intercept of a function is the point where the function crosses y-axis'.
At x=0, we have, [tex]y=0^2-2\times 0-8[/tex] i.e. y= -8
Therefore, the y-intercept is (0,-8)
And, 'the x-intercept of a function is the point where the function crosses x-axis'.
So, for y=0, we have [tex]0=x^2-2x-8[/tex] i.e. [tex]0=(x-4)(x+2)[/tex] i.e. x= 4 and x= -2.
Thus, the x-intercept are (4,0) and (-2,0).
