Use the definition of $f(x)$ and set $f(x)=0$ to find x-intercepts. Write out the equation $-2(x-3)^2+10=0$ and think about how to isolate $(x-3)^2$.

Once you have an equation like $(x-3)^2 = 5$, remember that there are usually two solutions. What do you get when you take the square root of both sides?

In Desmos, type y = -2(x - 3)^2 + 10 to graph the function.

Look for the points where the graph crosses the x-axis (where $y=0$). Click those intersection points; Desmos will display their coordinates. Note the x-coordinates of both points, and then choose the option that lists both of those x-values.

The x-intercepts are the points where the graph crosses the x-axis. On the x-axis, $y=0$, so for this function we must solve

That is, find all $x$ values that make $-2(x-3)^2+10=0$.

Start with the equation

Move 10 to the other side and then divide by $-2$:

Now you have a simple equation of the form $(\text{something})^2 = 5$.

To solve $(x-3)^2 = 5$, take the square root of both sides. Remember to include both the positive and negative roots:

This gives you two equations to solve:

• $x - 3 = \sqrt{5}$
• $x - 3 = -\sqrt{5}$.

Solve each linear equation:

• From $x - 3 = \sqrt{5}$, add 3: $x = 3 + \sqrt{5}$.
• From $x - 3 = -\sqrt{5}$, add 3: $x = 3 - \sqrt{5}$.

So there are two real x-intercepts, at $x = 3 - \sqrt{5}$ and $x = 3 + \sqrt{5}$, and the correct answer choice is the option that lists both of these values.