Question

The graph of h is a translation 3 units up and 2 units right of the graph of f(x)=x2+4x. For each value of x, g(x) is 130% of h(x). Write a rule for g.

Answer 1

We are given

$f(x)=x^2+4x$

h(x):

The graph of h is a translation 3 units up

so, we get

$=x^2+4x+3$

2 units right of the graph of f(x)

so, we get

$h(x)=(x-2)^2+4(x-2)+3$

now, we can simplify it

$h(x)=x^2-4x+4+4\left(x-2\right)+3$

$h(x)=x^2-1$

g(x):

we have

For each value of x, g(x) is 130% of h(x)

so, we get

g(x)=130% of h(x)

$g(x)=\frac{130}{100} h(x)$

we can plug value

$g(x)=\frac{130}{100} (x^2-1)$

$g(x)=1.3(x^2-1)$.............Answer

Answer 2

Answer:

The rule for g is:

                      $g(x)=1.3(x^2-1)$

Step-by-step explanation:

The equation of the function f(x) is given by:

$f(x)=x^2+4x$

Now, it is given that:

The graph of h is a translation 3 units up and 2 units right of the graph of f(x).

This means that:

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

Hence, we have the equation for the function h(x) as:

$h(x)=(x-2)^2+4(x-2)+3\\\\h(x)=x^2+4-4x+4x-8+3\\\\h(x)=x^2+4-8+3\\\\h(x)=x^2-1$

Also, g(x) is 130% of h(x).

i.e.

$g(x)=130\% \ of \ h(x)\\\\g(x)=1.3h(x)\\\\g(x)=1.3(x^2-1)$