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

View step-by-step help

Home

Math Problems

Algebra 1

Identify direct variation and inverse variation

Full solution

Q. Google Classroom

\newline

What is the maximum vertical distance between the line

y=x+2

and the parabola

y=x^{2}

for

-1 \leq x \leq 2

Understand the Problem: Understand the problem. $\newline$ We need to find the maximum vertical distance between the line $y = x + 2$ and the parabola $y = x^2$ within the interval $-1 \leq x \leq 2$ . The vertical distance at any point $x$ is given by the absolute value of the difference between the $y$ -values of the line and the parabola.
Set up the Equation: Set up the equation for the vertical distance. The vertical distance $D(x)$ between the line and the parabola at any point $x$ is $D(x) = |(x + 2) - x^2|$ .
Simplify the Equation: Simplify the equation for the vertical distance. $D(x) = |x + 2 - x^2| = |-(x^2 - x - 2)|$ .
Factor the Quadratic Expression: Factor the quadratic expression inside the absolute value. $\newline$ The quadratic $x^2 - x - 2$ can be factored as $(x - 2)(x + 1)$ . $\newline$ So, $D(x) = |-(x - 2)(x + 1)|$ .
Find Critical Points: Find the critical points of $D(x)$ within the interval $-1 \leq x \leq 2$ . The critical points will be the roots of the quadratic and any points where the quadratic changes sign. The roots are $x = 2$ and $x = -1$ . We also need to check the endpoints of the interval, which are $x = -1$ and $x = 2$ .
Evaluate at Critical Points: Evaluate $D(x)$ at the critical points and endpoints. $D(-1) = |-( -1 - 2)( -1 + 1)| = |-(3)(0)| = 0.$ $D(2) = |-(2 - 2)(2 + 1)| = |-(0)(3)| = 0.$ Since the roots give us a distance of $0$ , we need to check for the maximum distance at points between the roots.
Check for Maximum Distance: Since the maximum distance won't occur at the roots, we need to check within the interval. We can either use calculus to find the maximum or evaluate $D(x)$ at several points in the interval to estimate the maximum. $\newline$ Let's evaluate $D(x)$ at the midpoint of the interval, $x = 0.5$ , as an example. $\newline$ $D(0.5) = |-(0.5)^2 - 0.5 - 2| = |-(0.25 - 0.5 - 2)| = |-(-2.25)| = 2.25$ .
Take Derivative of $f(x)$ : To ensure we have the maximum, we can take the derivative of $D(x)$ with respect to $x$ and find where the derivative is zero. However, since $D(x)$ involves an absolute value, we need to consider the function without the absolute value and look for maximums on both the positive and negative intervals. $\newline$ Let's consider the function without the absolute value: $f(x) = -(x^2 - x - 2)$ .
Find Critical Point of Derivative: Take the derivative of $f(x)$ with respect to $x$ . $f'(x) = -2x + 1$ .
Verify and Evaluate at $x=\frac{1}{2}$ : Find the critical point where the derivative is zero. $0 = -2x + 1$ $2x = 1$ $x = \frac{1}{2}.$
Determine Maximum Vertical Distance: Verify that $x = \frac{1}{2}$ is within our interval and evaluate $D(x)$ at $x = \frac{1}{2}$ . We already evaluated $D(0.5)$ in Step $7$ and found it to be $2.25$ , which is within our interval.
Determine Maximum Vertical Distance: Verify that $x = \frac{1}{2}$ is within our interval and evaluate $D(x)$ at $x = \frac{1}{2}$ . We already evaluated $D(0.5)$ in Step $7$ and found it to be $2.25$ , which is within our interval.Since the derivative changes sign around $x = \frac{1}{2}$ , this indicates a maximum. We have already found that the vertical distance at $x = \frac{1}{2}$ is $2.25$ , which is the maximum vertical distance within the interval $-1 \leq x \leq 2$ .