listenconsider this system of equations.$\begin{cases} y=3x-3 \\ y=x^2-…

Question

listenconsider this system of equations.$\begin{cases} y=3x-3 \\ y=x^2-2x-3 end{cases}$plot these equations on the same graph.click the line button, then click on the two points to connect them in the graph.click the parabola button for the quadratic equation, then plot the vertex and the $y$-intercept.

Solution Steps

Understand the question
listenconsider this system of equations.$\begin{cases} y=3x-3 \\ y=x^2-2x-3 end{cases}$plot these equations on the same graph.click the line button, then click on the two points to connect them in the graph.click the parabola button for the quadratic equation, then plot the vertex and the $y$-intercept.
Explanation
Step1: Find line points

For $y=3x-3$:
When $x=0$, $y=3(0)-3=-3$ → point $(0, -3)$
When $x=1$, $y=3(1)-3=0$ → point $(1, 0)$

Step2: Find parabola vertex

For $y=x^2-2x-3$, use vertex formula $x=-\frac{b}{2a}$ where $a=1, b=-2$:
$x=-\frac{-2}{2(1)}=1$
Substitute $x=1$: $y=(1)^2-2(1)-3=1-2-3=-4$ → vertex $(1, -4)$

Step3: Find parabola y-intercept

For $y=x^2-2x-3$, set $x=0$:
$y=0^2-2(0)-3=-3$ → point $(0, -3)$

Step4: Find intersection points

Set $3x-3=x^2-2x-3$:
$x^2-5x=0$
$x(x-5)=0$
Solutions: $x=0$ (gives $y=-3$) and $x=5$ (gives $y=3(5)-3=12$) → points $(0, -3)$ and $(5, 12)$
Final answer
1. For the line $y=3x-3$: plot points $(0, -3)$ and $(1, 0)$, then connect them.
2. For the parabola $y=x^2-2x-3$: plot vertex $(1, -4)$ and y-intercept $(0, -3)$, then sketch the parabola (which also passes through the intersection point $(5, 12)$).
3. The system's solutions (intersection points) are $(0, -3)$ and $(5, 12)$.

Answer

Explanation

Step1: Find line points

For $y=3x-3$:
When $x=0$, $y=3(0)-3=-3$ → point $(0, -3)$
When $x=1$, $y=3(1)-3=0$ → point $(1, 0)$

Step2: Find parabola vertex

For $y=x^2-2x-3$, use vertex formula $x=-\frac{b}{2a}$ where $a=1, b=-2$:
$x=-\frac{-2}{2(1)}=1$
Substitute $x=1$: $y=(1)^2-2(1)-3=1-2-3=-4$ → vertex $(1, -4)$

Step3: Find parabola y-intercept

For $y=x^2-2x-3$, set $x=0$:
$y=0^2-2(0)-3=-3$ → point $(0, -3)$

Step4: Find intersection points

Set $3x-3=x^2-2x-3$:
$x^2-5x=0$
$x(x-5)=0$
Solutions: $x=0$ (gives $y=-3$) and $x=5$ (gives $y=3(5)-3=12$) → points $(0, -3)$ and $(5, 12)$

Answer

For the line $y=3x-3$: plot points $(0, -3)$ and $(1, 0)$, then connect them.
For the parabola $y=x^2-2x-3$: plot vertex $(1, -4)$ and y-intercept $(0, -3)$, then sketch the parabola (which also passes through the intersection point $(5, 12)$).
The system's solutions (intersection points) are $(0, -3)$ and $(5, 12)$.

Question Image

Click to enlarge

Question Analysis

Subject mathematics

Sub Subject calculus

Education Level high school

Difficulty unspecified

Question Type with image

Multi Question No

Question Count 1

Analysis Status completed

Analyzed At 2026-02-09T20:26:29

OCR Text

Show OCR extraction

listenconsider this system of equations.$\begin{cases} y=3x-3 \\ y=x^2-2x-3 end{cases}$plot these equations on the same graph.click the line button, then click on the two points to connect them in the graph.click the parabola button for the quadratic equation, then plot the vertex and the $y$-intercept.

Sovi AI iOS

Official website: mysovi.ai. Question pages are served on question-banks.mysovi.ai. iOS app is distributed through the Apple App Store.

Download on App Store Category: Calculus

Sovi AI