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What is the maximum vertical distance between the parabola
y=3-x^(2) and the line
y=x+1 for
-2 <= x <= 1 ?

What is the maximum vertical distance between the parabola $y=3-x^{2}$ and the line $y=x+1$ for $-2 \leq x \leq 1$ ?

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Characteristics of quadratic functions: equations

Full solution

Q. What is the maximum vertical distance between the parabola

y=3-x^{2}

and the line

y=x+1

for

-2 \leq x \leq 1

?

Find Parabola Vertex: First, let's find the vertex of the parabola $y=3-x^2$ , which is in the form $y=a-x^2$ where $a=3$ . The vertex of this parabola is at $(0,3)$ since it's in the form $y=a-x^2$ and the $x$ -coordinate of the vertex is $0$ .
Calculate Line Y-value: Now, let's find the y-value of the line $y=x+1$ at $x=0$ to compare it with the vertex of the parabola. $\newline$ Substitute $x=0$ into $y=x+1$ to get $y=0+1$ , which is $y=1$ .
Calculate Vertical Distance at $x=0$ : The vertical distance between the parabola and the line at $x=0$ is the absolute difference of their $y$ -values. $\newline$ So, the distance at $x=0$ is $|3 - 1| = 2$ .
Check Endpoints: We need to check the endpoints of the interval $[-2, 1]$ to ensure we find the maximum vertical distance. $\newline$ Let's substitute $x=-2$ into both equations to find their y-values. $\newline$ For the parabola, $y=3-(-2)^2=3-4=-1$ . $\newline$ For the line, $y=(-2)+1=-1$ .
Calculate Vertical Distance at $x=-2$ : The vertical distance between the parabola and the line at $x=-2$ is $|(-1) - (-1)| = 0$ .
Calculate Vertical Distance at $x=1$ : Now, let's substitute $x=1$ into both equations to find their $y$ -values. $\newline$ For the parabola, $y=3-(1)^2=3-1=2$ . $\newline$ For the line, $y=(1)+1=1+1=2$ .

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