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

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…

分類: calculus 更新時間: 2026-02-09

題目

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.

解題步驟

  1. 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.

  2. 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 $(5, 12)$

  3. 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 $(5, 12)$.

答案

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

  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 $(5, 12)$.

Question Image

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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.

相關知識點

mathematicscalculuswith imagehigh schoolturns-1

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