Question

1. if the point ((a, b)) is on the graph (y = x^2), which of the following points in terms of (a) and (b) are on the graph (y = -(x - 3)^2 + 1)?

a) ((-a - 3, b + 1))
b) ((-a + 3, b + 1))
c) ((a + 3, -b - 1))
d) ((a + 3, -b + 1))

Explanation:

Step1: Use the given point on \( y = x^2 \)

Since \((a, b)\) is on \( y = x^2 \), we have \( b = a^2 \).

Step2: Substitute into the new function

We need to check which point \((x, y)\) satisfies \( y = -(x - 3)^2 + 1 \). Let's substitute each option's \( x \) and \( y \) into the equation.
For option d: \( x = a + 3 \), \( y = -b + 1 \)
Substitute \( x \) into the right - hand side of \( y = -(x - 3)^2+1 \):
\[

$$\begin{align*} -( (a + 3)-3)^2+1&=-a^2 + 1\\ \end{align*}$$

\]
Since \( b=a^2 \), then \( -b + 1=-a^2 + 1 \), which matches the left - hand side \( y=-b + 1 \).
Let's check other options for completeness:

• Option a: \( x=-a - 3 \), \( y = b + 1 \)

\[

$$\begin{align*} -(-a - 3-3)^2+1&=-(-a - 6)^2+1=-(a^{2}+12a + 36)+1=-a^{2}-12a - 35 \end{align*}$$

\]
And \( b + 1=a^{2}+1\), which is not equal to \( -a^{2}-12a - 35 \) (unless \( a = 0\) and \( a=- 6\) at the same time, which is not general).

• Option b: \( x=-a + 3 \), \( y = b + 1 \)

\[

$$\begin{align*} -(-a + 3-3)^2+1&=-(-a)^2+1=-a^{2}+1 \end{align*}$$

\]
And \( y=b + 1=a^{2}+1\), \( -a^{2}+1
eq a^{2}+1 \) (unless \( a = 0\)).

• Option c: \( x=a + 3 \), \( y=-b-1 \)

\[

$$\begin{align*} -(a + 3-3)^2+1&=-a^{2}+1 \end{align*}$$

\]
And \( y=-b - 1=-a^{2}-1\), \( -a^{2}+1
eq -a^{2}-1 \)

Answer:

d) \((a + 3,-b + 1)\)