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x

y

21
-8

23
8

25
-8

The table shows three values of
x and their corresponding values of
y, where
y=f(x)+4 and
f is a quadratic function. What is the
y-coordinate of the
y-intercept of the graph of
y=f(x) in the
xy-plane?

\begin{tabular}{|c|c|} $\newline$ \hline $x$ & $y$ \\ $\newline$ \hline $21$ & $-8$ \\ $\newline$ \hline $23$ & $8$ \\ $\newline$ \hline $25$ & $-8$ \\ $\newline$ \hline $\newline$ \end{tabular} $\newline$ The table shows three values of $x$ and their corresponding values of $y$ , where $y=f(x)+4$ and $f$ is a quadratic function. What is the $y$ -coordinate of the $y$ -intercept of the graph of $y=f(x)$ in the $x y$ -plane?

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Reflections of functions

Full solution

Q. \begin{tabular}{|c|c|}

\newline

\hline

x

&

y

\\

\newline

\hline

21

&

-8

\\

\newline

\hline

23

&

8

\\

\newline

\hline

25

&

-8

\\

\newline

\hline

\newline

\end{tabular}

\newline

The table shows three values of

x

and their corresponding values of

y

, where

y=f(x)+4

and

f

is a quadratic function. What is the

y

-coordinate of the

y

-intercept of the graph of

y=f(x)

in the

x y

-plane?

Understand y-intercept calculation: First, we need to understand that the y-coordinate of the y-intercept of the graph of $y=f(x)$ is the value of $f(x)$ when $x=0$ . Since we are given that $y=f(x)+4$ , we can find the y-coordinate of the y-intercept by finding $f(0)$ and then subtracting $4$ .
Find coefficients using given points: We are given three points on the graph of $y=f(x)+4$ . We can use these points to find the coefficients of the quadratic function $f(x)=ax^2+bx+c$ . Since we know that $y=f(x)+4$ , we can rewrite the points for $f(x)$ by subtracting $4$ from the $y$ -values.
Adjust points for $f(x)$ : The adjusted points for $f(x)$ are as follows: $\newline$ $x=21$ , $f(x)=y-4=-8-4=-12$ $\newline$ $x=23$ , $f(x)=y-4=8-4=4$ $\newline$ $x=25$ , $f(x)=y-4=-8-4=-12$ $\newline$ Now we have three points $(21, -12)$ , $(23, 4)$ , and $f(x)$ $0$ that lie on the graph of $f(x)$ .
Set up system of equations: We can set up a system of equations using these points and the general form of a quadratic function $f(x)=ax^2+bx+c$ :
For $x=21$ , $f(21)=a(21)^2+b(21)+c=-12$
For $x=23$ , $f(23)=a(23)^2+b(23)+c=4$
For $x=25$ , $f(25)=a(25)^2+b(25)+c=-12$
Solve system of equations: Let's solve the system of equations. We'll start with the first and third equations to find a relationship between $a$ and $c$ because the $y$ -values are the same for $x=21$ and $x=25$ , which suggests that $b$ might cancel out. $\newline$ $a(21)^2+c=-12$ $\newline$ $a(25)^2+c=-12$ $\newline$ Subtracting the first equation from the second gives us: $\newline$ $a(25)^2 - a(21)^2 = 0$ $\newline$ $a(625 - 441) = 0$ $\newline$ $a(184) = 0$ $\newline$ This implies that $a=0$ , which cannot be true for a quadratic function. There must be a mistake in our calculations.

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