   Chapter 3.1, Problem 70E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find a parabola with equation y = ax2 + bx + c that has slope 4 at x = 1, slope –8 at x = –1, and passes through the po.int (2, 15).

To determine

To find: The equation of parabola.

Explanation

Given:

The equation of the parabola y=ax2+bx+c (1)

The slope of the parabola are 4 at x=1 and -8 at x=1.

The parabola passing through the point (2,15).

Derivative rules:

(1) Constant multiple rule: ddx(cf)=cddx(f)

(2) Power rule: ddx(xn)=nxn1

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

Calculation:

Obtain the values of a, b and c in the equation of parabola.

Take the point (2,15).

Here, x=2 and y=15 .

Substitute 2 for x and 15 for y in equation (1).

15=a(2)2+b(2)+c(2)15=a(4)+b(2)+c(2)15=4a+2b+c

Therefore, the equation is 4a+2b+c=15 (2)

Obtain the derivative of y=ax2+bx+c.

y(x)=ddx(ax2+bx+c)

Apply the sum rule (3) and the constant multiple rule (1),

Since the derivative of constant function is zero, then ddx(c)=0

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