Let f(x)=12x2+4.

Mathematics

2 answers:

Jlenok [28]3 years ago

6 0

Answer:

$g(x)= 48x^2+16$

Step-by-step explanation:

given f(x)= $12x^2 +4$

The function g(x) is a vertical expansion of f(x) by a factor of 4.

when f(x) is vertically expansion by 'a' f(x) becomes a f(x)

For vertical expansion by a factor of 4 we multiply f(x) by 4

So $g(x)= 4(f(x))$

$g(x)= 4(12x^2+4)$

$g(x)= 48x^2+16$

PtichkaEL [24]3 years ago

4 0

G(X)=
16x^2+4
Parent function is y=x^2

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Please help find domain and range

Oxana [17]

<h3>Answer: Choice A</h3>

Domain: x > 4
Range: y > 0

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Explanation:

We want to avoid having a negative number under the square root. Solving $x-4 \ge 0$ leads to $x \ge 4$

So it appears the domain could involve x = 4 itself; however, if we tried that x value, then we'd get a division by zero error.

So in reality, the domain is x > 4.

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The range of y = sqrt(x) is the set of positive real numbers. So y > 0 is the range for this equation. Shifting left and right does not affect the range, so the range of y = sqrt(x-4) is also y > 0.

We are dividing a positive number (3) over some positive number in the denominator. Overall, the expression $\frac{3}{\sqrt{x-4}}$ is positive because positive/positive = positive.

Therefore, the range of the given equation is y > 0

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The graph is shown below. We have a vertical asymptote at x = 4 and a horizontal asymptote at y = 0. The green curve is fenced in the upper right corner (northeast corner).