Question

Rewrite the function in standard form, intercept form, find the vertex, find the y-intercept, and find the x-intercepts.

Answer 1

Answer:

We have the function, $f(x)=(x+3)^{2}-4$

On simplifying, we get,

$f(x)=x^2+6x+9-4$

i.e. $f(x)=x^2+6x+5$

Thus, the standard form of the function is $f(x)=x^2+6x+5$.

Now, the factors of the given functions are (x+1) and (x+5).

Since, the intercept form of the function is the factored form of the function.

So, we have, intercept form of the function is $f(x)=(x+1)(x+5)$.

Now, we know that,

Value of x-coordinate of the vertex is $\frac{-b}{2a}$ i.e. $\frac{-6}{2\times 1}$ i.e. $\frac{-6}{2}$ i.e. x= -3

Then, $f(-3)=(-3)^2+6\times (-3)+5$ i.e. $f(-3)=9-18+5$ i.e. f(-3)=-4

So, the vertex of the function is (-3,-4).

Further, we know that, 'the y-intercept of a function is the point where the function crosses y-axis'.

So, when x=0, we have, $f(0)=0^2+6\times 0+5$ i.e. f(0) = 5

Thus, the y-intercept is (0,5)

Also, 'the x-intercept of a function is the point where the function crossese x-axis'.

Then, for f(x)=0, we have $0=x^2+6x+5$ i.e. $0=(x+1)(x+5)$ i.e. x= -1 and x= -5

Thus, the x-intercept are (-1,0) and (-5,0).