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

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Q. What is the maximum vertical distance between the line

y=x+2

and the parabola

y=x^{2}

for

-1 \leq x \leq 2

Calculate Difference Function: To find the maximum vertical distance, we need to calculate the difference between the y-values of the line and the parabola for any $x$ in the given interval. $\newline$ Difference function: $d(x) = |(x + 2) - x^2|$
Simplify Difference Function: Simplify the difference function to make it easier to work with. $\newline$ $d(x) = |x + 2 - x^2| = |-x^2 + x + 2|$
Find Vertex of Parabola: To find the maximum value of $d(x)$ , we need to find the vertex of the parabola represented by $d(x)$ , since the vertex will give us the maximum or minimum value of the function. $\newline$ The vertex form of a parabola is $y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex.
Rewrite in Vertex Form: First, we need to rewrite $d(x)$ in vertex form. To do this, we complete the square. $d(x) = -(x^2 - x) + 2$ Add and subtract $(1/2)^2$ inside the parenthesis to complete the square. $d(x) = -(x^2 - x + (1/2)^2) + 2 + (1/2)^2$
Identify Parabola Vertex: Simplify the equation by combining like terms and factoring. $\newline$ $d(x) = -(x - \frac{1}{2})^2 + 2 + \frac{1}{4}$ $\newline$ $d(x) = -(x - \frac{1}{2})^2 + \frac{9}{4}$
Find Maximum Value: Now that we have the vertex form, we can identify the vertex of the parabola. The vertex is at $(h, k) = (\frac{1}{2}, \frac{9}{4})$ . Since the coefficient of the squared term is negative, the parabola opens downwards, and the vertex represents the maximum point.
Check Vertex Interval: The maximum value of $d(x)$ is the y-coordinate of the vertex, which is $\frac{9}{4}$ . This is the maximum vertical distance between the line and the parabola within the given interval.
Final Maximum Vertical Distance: However, we need to check if the vertex falls within the interval $-1 \leq x \leq 2$ . Since $h = \frac{1}{2}$ , it does fall within the interval.
Final Maximum Vertical Distance: However, we need to check if the vertex falls within the interval $-1 \leq x \leq 2$ . Since $h = \frac{1}{2}$ , it does fall within the interval.Therefore, the maximum vertical distance between the line $y=x+2$ and the parabola $y=x^2$ for $-1 \leq x \leq 2$ is $\frac{9}{4}$ .