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

Question

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

Bytelearn · Accepted Answer

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. 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. 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. Let's substitute x=-2 into both equations to find their y-values. For the parabola, y=3-(-2)^2=3-4=-1. 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. For the parabola, y=3-(1)^2=3-1=2. For the line, y=(1)+1=1+1=2.

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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What is the maximum vertical distance between the parabola y=3−x2 y=3-x^{2} y=3−x2 and the line y=x+1 y=x+1 y=x+1 for −2≤x≤1 -2 \leq x \leq 1 −2≤x≤1 ?

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