Question 98·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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The system of equations is

y = x^2 + 4x + 1

y = x + 5

When the equations above are graphed in the $xy$ -plane, what are the coordinates $(x, y)$ of the points of intersection of the two graphs?

A linear-nonlinear system pairs a line with a curve; a squared $x$ tells you the curve is a parabola. Type both equations into Desmos and click every intersection, since each shared point satisfies both equations. Report the points in $(x,y)$ order. Swapping the numbers within a pair puts the point somewhere else.

Hints

Hint 1
The squared term makes one graph a parabola, while $y=x+\text{constant}$ makes a line. A solution must lie on both graphs. What do the places where the graphs cross tell you?
Hint 2
At each crossing, Desmos shows an ordered pair: the first number is $x$ , and the second is $y$ . Click each crossing rather than stopping at one, then keep the two numbers in that order.

Step-by-step

Approach 1: Graph and click both intersections
Step 1
Find where the graphs cross
Type $y=x^2+4x+1$ and $y=x+5$ on separate Desmos lines. The squared equation graphs as a curved parabola; the other graphs as a line. Desmos shows two crossings, so there are two points to find. Each crossing lies on both graphs.
Step 2
Read both points in order
Click each crossing. Desmos labels them $(-4,1)$ and $(1,6)$ . In $(x,y)$ , $x$ comes first, so don't reverse either pair. Keep each intersection's $x$ - and $y$ -coordinates together. The points of intersection are $(-4,1)$ and $(1,6)$ . Choice D.

Approach 2: Set the expressions equal and factor
Step 1
Make the two values of y equal
At an intersection, both equations give the same $y$ . Set their right sides equal, a move called substitution:
$x^2+4x+1=x+5.$
Step 2
Put zero on one side
Subtract $x+5$ from both sides to get zero on one side:
$x^2+3x-4=0.$
Step 3
Factor the quadratic
The numbers $4$ and $-1$ multiply to $-4$ and add to $3$ , so factor:
$(x+4)(x-1)=0.$
Step 4
Set each factor to zero
Use the zero-product property: if a product is zero, at least one factor is zero. So:
$x+4=0\quad\text{or}\quad x-1=0.$
Step 5
Find both x-values
Solve both short equations:
$\begin{aligned}x+4=0&\quad\Rightarrow\quad x=-4,\x-1=0&\quad\Rightarrow\quad x=1.\end{aligned}$
Step 6
Pair each x-value with its y-value
Use the line $y=x+5$ to find a separate $y$ -value for each $x$ . Substitute $x=-4$ : $y=-4+5=1$ . Substitute $x=1$ : $y=1+5=6$ . So the points of intersection are $(-4,1)$ and $(1,6)$ . Choice D.

Question 98·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

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