Question 146·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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A parabola $P$ and a line $L$ are graphed in the $xy$ -plane. The parabola has labeled points $A$ and $B$ on the $x$ -axis and a labeled vertex $V$ . The line has labeled points $C$ and $D$ .

The solutions to the system are the points where $P$ and $L$ intersect. Which choice is the $x$ -coordinate of the solution with $y>0$ ?

A system of equations with a line and a parabola can have two intersections. If a figure gives the parabola’s intercepts and vertex instead of an equation, use those points to build its equation; use two points to build the line’s equation, then graph both in Desmos. Use the given condition to choose an intersection before reading the requested coordinate. A point on the $x$ -axis does not satisfy $y>0$ .

Hints

Hint 1
At an $x$ -intercept, $y=0$ . Each of the parabola’s two intercepts gives an $x$ -value that makes one factor of its equation zero. What factors do those values give you?
Hint 2
A vertex is the parabola’s turning point. Put its coordinates into your equation to find the multiplier in front of the factors; the intercepts alone don’t determine the parabola’s height.
Hint 3
The line’s slope is its change in $y$ divided by its change in $x$ . Find that slope from the labeled points, then graph both equations and choose the intersection with positive $y$ .

Step-by-step

Build both equations and graph their intersections
Step 1
Use the parabola’s intercepts
An $x$ -intercept has $y=0$ . The figure shows $A(-1,0)$ and $B(5,0)$ , so $x+1$ and $x-5$ must each make the parabola’s equation zero. Write
$y=a(x+1)(x-5).$
The unknown multiplier $a$ controls the parabola’s height.
Step 2
Use the vertex to make an equation
The vertex $V(2,9)$ lies on the parabola, so put its coordinates into the equation:
$9=a(2+1)(2-5).$
Using the vertex matters: the intercepts alone cannot tell you $a$ .
Step 3
Find the multiplier
Type $9\sim a(2+1)(2-5)$ in Desmos. The $\sim$ tells Desmos to fit the unknown multiplier $a$ ; it reports $a=-1$ under PARAMETERS.
Step 4
Find the line’s slope
The slope is the change in $y$ divided by the change in $x$ . From $C(1,3)$ to $D(5,9)$ , type $\frac{9-3}{5-1}$ in Desmos. It shows $1.5$ ; its fraction button gives $\frac32$ . So the line rises $3$ for every $2$ units right.
Step 5
Write and graph the line
Start at $C(1,3)$ and use the slope $\frac32$ : the rise from $3$ is $\frac32$ times the run from $1$ . In point-slope form, type
$y-3=\frac32(x-1).$
Desmos draws the line through $C$ and $D$ .
Step 6
Graph the parabola
Using $a=-1$ , type the parabola as $y=-(x+1)(x-5)$ below the line equation. Desmos draws the parabola through $A$ and $B$ , crossing the line once on the $x$ -axis and once above it.
Step 7
Choose the intersection above the axis
Click both intersections: Desmos shows $(-1,0)$ and $(3.5,6.75)$ . The first has $y=0$ , so it fails $y>0$ . The second has positive $y$ , and its requested $x$ -coordinate is $3.5=\frac72$ . Choice B.

Question 146·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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