Question

seat ap precalculus test 1 - 2 answer questions 10 and 11 about the graph of y = f(x) depicted in the graph below: 10. which combination of descriptions applies to f(x)? (a) odd, one - to - one, and a function (b) one - to - one, a function, but not odd (c) odd, a function but not one - to - one (d) not odd, not one - to - one, but a function (e) not odd, not one - to - one, and not a function 11. which point will be present on the graph of y=-f(2x)? (a) (1,2) (b) (4,2) (c) (-1,2) (d) (1,-2) (e) (-4,-2) - the reflection makes it positive 12. consider a pair of transformations that transform y = x² - 8x - 3 to y = x² + 8x + 3. what would those transformations be? (a) reflection across y - axis and shift 6 units up (b) reflection across x - axis and shift 6 units up (c) reflection across the y - axis and shift 6 units left (d) reflection across the x - axis and shift 6 units left (e) none of these - negate x - add 6

Explanation:

10.

Step1: Check if it's a function

Use the vertical - line test. Since any vertical line drawn on the graph intersects the curve at most once, it is a function.

Step2: Check one - to - one

A function is one - to - one if any horizontal line intersects the graph at most once. A horizontal line \(y = 3\) (for example) intersects the graph at more than one point, so it is not one - to - one.

Step3: Check if it's odd

For a function \(y = f(x)\) to be odd, \(f(-x)=-f(x)\) for all \(x\) in the domain. The graph is not symmetric about the origin, so it is not odd.

Step1: Recall transformation rules

For \(y=-f(2x)\), first, the horizontal compression by a factor of \(\frac{1}{2}\) and then a reflection about the \(x\) - axis. Let's assume a point \((x_0,y_0)\) on \(y = f(x)\) transforms to \((x_1,y_1)\) on \(y=-f(2x)\). The horizontal transformation is \(x_1=\frac{x_0}{2}\), and the vertical transformation is \(y_1=-y_0\).

Step2: Test points

Let's assume a point \((2, - 2)\) on \(y = f(x)\) (by looking at the general shape of the graph). For \(y=-f(2x)\), when \(x_0 = 2\), \(x_1=\frac{2}{2}=1\), and \(y_0=-2\), then \(y_1=-(-2)=2\). So the point \((1,2)\) is on the graph of \(y=-f(2x)\).

Step1: Analyze the \(x\) - terms

For the function \(y = x^{2}-8x - 3\) to \(y=x^{2}+8x + 3\), replacing \(x\) with \(-x\) in \(y=x^{2}-8x - 3\) gives \(y=(-x)^{2}-8(-x)-3=x^{2}+8x - 3\).

Step2: Analyze the constant term

To get from \(y=x^{2}+8x - 3\) to \(y=x^{2}+8x + 3\), we need to shift the graph 6 units up (\(y_{new}=y_{old}+6\)). So the transformation is a reflection across the \(y\) - axis and a shift 6 units up.

Answer:

D. not odd, not one - to - one, but a function

11.