A line and a parabola intersect where that quadratic equals 0. Exactly one intersection means the quadratic has exactly one real solution.

For $ax^2+bx+c=0$, exactly one real solution happens when $b^2-4ac=0$. Apply that to your quadratic to restrict $p$, then use the repeated-root formula $x=-\frac{b}{2a}$.

Enter the parabola:

y = -2x^2 + 7x - 50

Enter the line using a parameter:

y = p x - 32

Desmos will create a slider for $p$.

Move the slider for $p$ until the line just touches the parabola (the graphs meet at a single point rather than crossing at two points). Тhis quеѕtiоn‌ is ⁠frо‌m⁠ Аnikо

Click the intersection point and note its $x$-coordinate. (There are two different slider positions where the line is tangent.) Choose the $x$-value you see that matches one of the answer choices.

At an intersection point, the $y$-values are equal:

Move all terms to one side:

Multiply by $-1$ to make the leading coefficient positive:

A quadratic $ax^2+bx+c=0$ has exactly one real solution when its discriminant is 0:

Here, $a=2$, $b=p-7$, and $c=18$, so

From $(p-7)^2=144$, we have

so anі‌к​o.аi ЅAT Q‌uеѕtіоn B⁠аnk

When the discriminant is 0, the quadratic’s (repeated) solution is

Compute $x=-\frac{p-7}{4}$ for each allowed $p$:

• If $p=19$, then $x=-\frac{19-7}{4}=-\frac{12}{4}=-3$.
• If $p=-5$, then $x=-\frac{-5-7}{4}=-\frac{-12}{4}=3$.

So $x$ can be $-3$ or $3$ depending on $p$. Among the answer choices, the value that could be $x$ is -3.