Question

What would the value of a be in the function d(x)=a2x−3−−−−−√+1
to vertically stretch the function f(x)=2x−3−−−−−√+1
by a factor of 3/2
? Enter your response in fraction form if needed.(1 point)
The value of a would be
.

Answer 1

To vertically stretch the function \( f(x) = 2\sqrt{x - 3} + 1 \) by a factor of \( \frac{3}{2} \), we need to find the new function \( d(x) \).

When stretching a function vertically by a factor of \( k \), we multiply the entire function by \( k \). Therefore, the new function \( d(x) \) can be expressed as:

\[
d(x) = \frac{3}{2} f(x) = \frac{3}{2} (2\sqrt{x - 3} + 1)
\]

By simplifying this, we have:

\[
d(x) = \frac{3}{2} \cdot 2\sqrt{x - 3} + \frac{3}{2} \cdot 1 = 3\sqrt{x - 3} + \frac{3}{2}
\]

Now, the given form of the function \( d(x) \) is \( d(x) = a \cdot 2\sqrt{x - 3} + 1 \). To match it, we need to relate this form to the newly derived expression:

\[
3\sqrt{x - 3} + \frac{3}{2}
\]

To make the factors match, we can look at the coefficients in front of \( \sqrt{x - 3} \):

If we set \( a \cdot 2 = 3 \) (the coefficient of the square root in our new function), we find \( a \) as follows:

\[
a = \frac{3}{2}
\]

Thus, the value of \( a \) that results in a vertical stretch of the function \( f(x) \) by a factor of \( \frac{3}{2} \) is

\[
\boxed{\frac{3}{2}}
\]