Question

solve the system of equations by graphing. first graph the equations, and then identify the solutions.

$y = -2x - 3$

$y = (x + 3)^2 + 4$

to graph a linear equation, click to plot points on the graph. to graph a quadratic equation, click to plot the vertex first, then plot another point on the graph.

graph with x-axis from -10 to 10 and y-axis from -10 to 10, grid lines, and axes labeled x and y

Explanation:

Step1: Graph \( y = -2x - 3 \)

This is a linear equation in slope - intercept form \( y=mx + b \), where the slope \( m=-2 \) and the y - intercept \( b=-3 \). To graph it, we can find two points. When \( x = 0 \), \( y=-2(0)-3=-3 \), so we have the point \( (0,-3) \). When \( y = 0 \), \( 0=-2x - 3\), then \( 2x=-3 \), \( x=-\frac{3}{2}=-1.5 \), so we have the point \( (-1.5,0) \). We can plot these two points and draw a straight line through them.

Step2: Graph \( y=(x + 3)^{2}+4 \)

This is a quadratic equation in vertex form \( y=a(x - h)^{2}+k \), where the vertex is at \( (h,k)=(-3,4) \). To find another point, we can choose a value of \( x \) and find \( y \). Let's take \( x=-2 \), then \( y=(-2 + 3)^{2}+4=(1)^{2}+4=5 \), so we have the point \( (-2,5) \). We plot the vertex \( (-3,4) \) and the point \( (-2,5) \) (and we can find more points if needed) and draw the parabola.

Step3: Find the intersection points

After graphing both the line and the parabola, we observe that the line \( y=-2x - 3 \) and the parabola \( y=(x + 3)^{2}+4 \) do not intersect. We can also verify this algebraically by setting the two equations equal to each other:
\( -2x-3=(x + 3)^{2}+4 \)
Expand the right - hand side: \( -2x-3=x^{2}+6x + 9+4 \)
Simplify to get a quadratic equation: \( x^{2}+6x + 13+2x + 3=0 \)
\( x^{2}+8x+16 = 0 \)
The discriminant of the quadratic equation \( ax^{2}+bx + c = 0 \) is \( \Delta=b^{2}-4ac \). Here, \( a = 1 \), \( b = 8 \), \( c = 16 \), so \( \Delta=8^{2}-4\times1\times16=64 - 64=0 \)? Wait, no, wait:
Wait, when we set \( -2x-3=(x + 3)^{2}+4 \)
\( (x + 3)^{2}+2x+7 = 0 \)
\( x^{2}+6x + 9+2x+7=0 \)
\( x^{2}+8x + 16=0 \)
\( (x + 4)^{2}=0 \)
\( x=-4 \)
When \( x=-4 \), \( y=-2(-4)-3=8 - 3 = 5 \) and \( y=(-4 + 3)^{2}+4=(-1)^{2}+4=5 \)
Wait, I made a mistake in the earlier graphing thought. Let's re - evaluate.
If \( x=-4 \):
For \( y=-2x-3 \), \( y=-2\times(-4)-3 = 8 - 3=5 \)
For \( y=(x + 3)^{2}+4 \), \( y=(-4 + 3)^{2}+4=(-1)^{2}+4 = 1 + 4=5 \)
Let's check \( x=-5 \):
For \( y=-2x-3 \), \( y=-2\times(-5)-3=10 - 3 = 7 \)
For \( y=(x + 3)^{2}+4 \), \( y=(-5 + 3)^{2}+4=(-2)^{2}+4=4 + 4 = 8 \)
Wait, when \( x=-3 \):
For \( y=-2x-3 \), \( y=-2\times(-3)-3=6 - 3 = 3 \)
For \( y=(x + 3)^{2}+4 \), \( y=0 + 4=4 \)
Wait, let's solve \( -2x-3=(x + 3)^{2}+4 \) correctly:
\( (x + 3)^{2}+4=-2x-3 \)
\( x^{2}+6x + 9+4+2x + 3 = 0 \)
\( x^{2}+8x+16 = 0 \)
\( (x + 4)^{2}=0 \)
So \( x=-4 \) (a repeated root). So the system has one solution (a tangent point) at \( x=-4 \), \( y = 5 \). Wait, my initial graphing analysis was wrong. Let's re - graph:

For the line \( y=-2x - 3 \), when \( x=-4 \), \( y=-2\times(-4)-3=5 \)
For the parabola \( y=(x + 3)^{2}+4 \), when \( x=-4 \), \( y=(-4 + 3)^{2}+4=1 + 4=5 \)
Let's find another point on the line: when \( x=-3 \), \( y=-2\times(-3)-3=3 \)
On the parabola, when \( x=-3 \), \( y=4 \)
When \( x=-5 \), line: \( y=-2\times(-5)-3 = 7 \), parabola: \( y=(-5 + 3)^{2}+4=8 \)
When \( x=-2 \), line: \( y=-2\times(-2)-3 = 1 \), parabola: \( y=(-2 + 3)^{2}+4=5 \)

So the two graphs intersect at the point \( (-4,5) \) (since the quadratic equation \( x^{2}+8x + 16 = 0 \) has a repeated root \( x=-4 \), which means the line is tangent to the parabola at \( x=-4 \), \( y = 5 \))

Answer:

The solution of the system is \( x=-4,y = 5 \) (or the ordered pair \( (-4,5) \))