Question

The graph of quadratic function f(x) has a minimum at (-2,-3) and passes through the point (2,13). The function g(x) is represented by the equation
g(x) = -(x+2)(x-3).

How much greater is the y-intercept of g(x) than f(x)?

Answer 1

Answer:

The y-intercept of g(x) is 5 greater than the y-intercept of f(x).

Step-by-step explanation:

It is given that the graph of quadratic function f(x) has a minimum at (-2,-3) and passes through the point (2,13).

The equation of a quadratic function is

$f(x)=a(x-h)^2+k$

Where, (h,k) is vertex or extreme points and a is stretch factor.

The minimum value of function is (-2,-3), so the vertex is (-2,-3).

$f(x)=a(x-(-2))^2+(-3)$

$f(x)=a(x+2)^2-3$

It is given that the function passing through the point (2,13).

$13=a(2+2)^2-3$

$16=16a$

$a=1$

So, the function f(x) is

$f(x)=(x+2)^2-3$

Substitute x=0, to find the y-intercept.

$f(x)=(0+2)^2-3$

$f(x)=4-3=1$

The y-intercept of f(x) is 1.

The given function is

$g(x)=-(x+2)(x-3)$

Substitute x=0, to find the y-intercept.

$g(x)=-(0+2)(0-3)=6$

The y-intercept of g(x) is 6.

The difference between y-intercepts is

$6-1=5$

Therefore y-intercept of g(x) is 5 greater than the y-intercept of f(x).

Answer 2

ANSWER

The y-intercept of g(x) is 5 greater than f(x)?

EXPLANATION
The function f(x) has a minimum at (-2,-3) and passes through (2,13).
The equation in vertex form is given by,

$f(x)=a {(x - h)}^{2} + k$

$f(x)=a {(x + 2)}^{2} - 3$

We substitute (2,13) to find the value of a,

$13=a {(2 + 2)}^{2} - 3$

$a {(4)}^{2} = 16$

$a = \frac{16}{16}$

$a=1$

$f(x)={(x + 2)}^{2}-3$

The y-intercept is

$f(0)={( 2)}^{2} - 3 =1$

Also,

$g(x) = - (x + 2)(x - 3)$

has y-intercept,

$g(0) = - (0+ 2)(0 - 3) = 6$

The difference is

$6 -1= 5$