After you find $f(21)$, $f(23)$, and $f(25)$, compare them. For a quadratic function, if two $x$-values are the same distance from some center and their $y$-values match, what does that say about the axis of symmetry and the vertex?

Once you know the vertex $(h,k)$ of $f(x)$, write $f(x)$ in the form $a(x-h)^2 + k$ and plug in one of your known points to solve for $a$. Then use this formula to evaluate $f(0)$ for the $y$-intercept.

The $y$-intercept happens where the graph crosses the $y$-axis. Which $x$-value does that correspond to, and how can you use your formula for $f(x)$ to find the corresponding $y$-value?

Create a table in Desmos with the three points from the question: in the first column, enter $21$, $23$, and $25$; in the second column, enter $-8$, $8$, and $-8$. Desmos will label these as $x_1$ and $y_1$ by default.

In a new expression line, type y1 ~ ax^2 + bx + c. Desmos will perform a quadratic regression and show values for $a$, $b$, and $c$. This gives an equation for the function $y = f(x) + 4$ that matches the table.

In another expression line, define a new function using the regression equation but subtract 4: for example, if the regression found $y = ax^2 + bx + c$, type f(x) = ax^2 + bx + c - 4. This new function $f(x)$ represents the original quadratic the question is asking about.

In a final expression line, type f(0). The value Desmos displays is the $y$-coordinate of the $y$-intercept of the graph of $y = f(x)$.

The table is for the function $y = f(x) + 4$, but the question asks about the graph of $y = f(x)$.

Use the relationship $y = f(x) + 4$ to find $f(x)$ at the three $x$-values by subtracting 4 from each $y$ in the table:

• When $x = 21$: $f(21) + 4 = -8 \Rightarrow f(21) = -8 - 4 = -12$.
• When $x = 23$: $f(23) + 4 = 8 \Rightarrow f(23) = 8 - 4 = 4$.
• When $x = 25$: $f(25) + 4 = -8 \Rightarrow f(25) = -8 - 4 = -12$.

So the graph of $y = f(x)$ passes through the points $(21,-12)$, $(23,4)$, and $(25,-12)$.

A quadratic function’s graph is a parabola, which is symmetric about a vertical line called its axis of symmetry.

Notice that:

• $(21,-12)$ and $(25,-12)$ have the same $y$-value and are equally spaced around $x = 23$.

That means the axis of symmetry is $x = 23$, and the point on the parabola that lies on this axis between them is the vertex. The middle point $(23,4)$ is therefore the vertex of $f(x)$.

So the vertex of $f(x)$ is $(23,4)$, and the parabola opens downward (because the vertex has the highest $y$ among these three points).

The vertex form of a quadratic with vertex $(h,k)$ is

Here, $h = 23$ and $k = 4$, so

Use one of the known points, such as $(21,-12)$, to solve for $a$.

Substitute $x = 21$ and $f(21) = -12$:

• $-12 = a(21 - 23)^2 + 4$
• $-12 = a(-2)^2 + 4$
• $-12 = 4a + 4$
• $4a = -16$
• $a = -4$.

So the quadratic is

The $y$-intercept is the point where the graph crosses the $y$-axis, which occurs at $x = 0$.

Use the formula for $f(x)$ and plug in $x = 0$:

So the $y$-coordinate of the $y$-intercept of the graph of $y = f(x)$ is $-2112$.