A parabola and a line are graphed in the -plane. The parabola has labeled points and on the -axis and a labeled vertex . The line has labeled points and . The solutions to the system are the points where and intersect. Which choice is the -coordinate of the solution with ?

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SAT Nonlinear Equations & Systems Question 146 | Free SAT Prep | Aniko

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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 $y>0$ ?

When a graph shows a line and a parabola with key points labeled, use those points to build each equation efficiently: intercept form for the parabola (then use the vertex to find the scale factor) and slope-intercept form for the line (from two points). After solving the resulting system, do not forget to apply any extra condition given in the question (here, choosing the intersection with $y>0$ ).

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Use the labeled intercepts

Use points $A$ and $B$ to write the parabola in intercept form: $y=a(x-r_1)(x-r_2)$ .

Use the labeled vertex to find $a$

Substitute the coordinates of $V$ into your parabola equation to determine the value of $a$ .

Solve and then check the condition $y>0$

After you find the two intersection $x$ -values, use the graph (or either equation) to decide which one gives $y>0$ .

Desmos Guide

Enter the parabola from the graph’s features

Using the intercepts and vertex shown on the graph, enter the parabola as

y=-(x+1)(x-5).

Enter the line from two plotted points

Using points $C$ and $D$ , enter the line as

y=\frac{3}{2}x+\frac{3}{2}.

Find the intersection with positive y

Click the intersection points of the two graphs. Identify the intersection whose $y$ -coordinate is above the $x$ -axis, and take its $x$ -coordinate.

Step-by-step Explanation

Write an equation for the parabola

From the graph, the parabola has $x$ -intercepts at $A(-1,0)$ and $B(5,0)$ , so

P: \ y=a(x+1)(x-5).

It also has vertex $V(2,9)$ . Substitute $(2,9)$ :

9=a(2+1)(2-5)=a(3)(-3)=-9a \Rightarrow a=-1.

P: \ y=-(x+1)(x-5).

Write an equation for the line

The line passes through $C(1,3)$ and $D(5,9)$ . Its slope is

m=\frac{9-3}{5-1}=\frac{6}{4}=\frac{3}{2}.

Using point $C(1,3)$ in $y=mx+b$ :

3=\frac{3}{2}(1)+b \Rightarrow b=\frac{3}{2}.

L: \ y=\frac{3}{2}x+\frac{3}{2}.

Solve the system and select the intersection with $y>0$

Set the expressions for $y$ equal:

-(x+1)(x-5)=\frac{3}{2}x+\frac{3}{2}.

Expand the left side: $-(x^2-4x-5)=-x^2+4x+5$ . Then

-x^2+4x+5=\frac{3}{2}x+\frac{3}{2}.

Multiply by 2 to clear fractions:

-2x^2+8x+10=3x+3.

Bring all terms to one side:

-2x^2+5x+7=0 \quad\Rightarrow\quad 2x^2-5x-7=0.

Factor:

2x^2-5x-7=(2x-7)(x+1)=0.

So $x=-1$ or $x=\frac{7}{2}$ . The point with $x=-1$ is on the $x$ -axis (so $y=0$ , not $y>0$ ), so the solution with $y>0$ has $x=\frac{7}{2}$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation