QUESTION IMAGE
Question
- jared graphed three points to create d(x). he then multiplied all of the output values by a constant value to create a(x).
| x | d(x) |
| 1 | 16 |
| 3 | 10 |
| 5 | -8 |
| x | a(x) |
| 1 | 8 |
| 3 | 5 |
| 5 | -4 |
a. what factor did he multiply the output values of d(x) by to get a(x)?
b. did he create a vertical stretch or vertical shrink? how do you know?
describe how each new function is related to the given parent function.
- parent function: f(x)=|x|
new function: w(x)=7|x|
- parent function: f(x)=√x
new function: m(x)=-√(1/4)x
- parent function: f(x)=x²
new function: p(x)=-1.5x²
- parent function: f(x)=x
new function: d(x)=0.4x
describe all transformations that occur on the function f(x).
- 3f(x)+4
- (2/3)f(x - 2)
- f(1.25x)-7
- (7/6)f(-3 + x)
Step1: Find the multiplication factor
For \(x = 1\), \(\frac{a(1)}{d(1)}=\frac{8}{16}=\frac{1}{2}\). For \(x = 3\), \(\frac{a(3)}{d(3)}=\frac{5}{10}=\frac{1}{2}\). For \(x = 5\), \(\frac{a(5)}{d(5)}=\frac{- 4}{-8}=\frac{1}{2}\).
Step2: Determine the type of transformation
If the factor \(k\) by which the output - values are multiplied satisfies \(03.a
The new function \(w(x)=7|x|\) is related to the parent function \(f(x) = |x|\) by a vertical stretch. When we have a function \(y = kf(x)\) and \(k>1\), in this case \(k = 7\), the graph of \(y = f(x)\) is vertically stretched by a factor of \(k\).
For the parent function \(f(x)=\sqrt{x}\) and the new function \(m(x)=-\sqrt{\frac{1}{4}x}\), first, there is a horizontal stretch by a factor of 4 (because of the \(\frac{1}{4}x\) inside the square - root. When we have \(y = f(bx)\) and \(0 < b<1\), the graph is horizontally stretched by a factor of \(\frac{1}{b}\), here \(b=\frac{1}{4}\) so the stretch factor is 4). Also, there is a reflection about the \(x\) - axis due to the negative sign in front of the square - root.
For the parent function \(f(x)=x^{2}\) and the new function \(p(x)=-1.5x^{2}\), there is a vertical stretch by a factor of 1.5 (since \(|k| = 1.5>1\) in \(y = kf(x)\)) and a reflection about the \(x\) - axis due to the negative sign.
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\(\frac{1}{2}\)