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QuestionFind the secant line between x=−3x=-3 and x=−1x=-1 for y=f(x)=x2+xy=f(x)=x^{2}+x and the tangent line at x=−3x=-3.

Studdy Solution

STEP 1

Assumptions1. The function is y=f(x)=x+xy=f(x)=x^{}+x . The two given x-values are x=−3x=-3 and x=−1x=-1
3. We are asked to find the equation of the secant line through the points where xx has the given values4. We are also asked to find the equation of the tangent line when xx has the first value

STEP 2

First, we need to find the y-values corresponding to the given x-values. We can do this by substituting the x-values into the function.
y=f(x)=x2+xy=f(x)=x^{2}+x

STEP 3

Substitute x=−3x=-3 into the function to find the corresponding y-value.
y=f(−3)=(−3)2+(−3)y=f(-3)=(-3)^{2}+(-3)

STEP 4

Calculate the y-value when x=−3x=-3.
y=f(−3)=9−3=6y=f(-3)=9-3=6

STEP 5

Substitute x=−1x=-1 into the function to find the corresponding y-value.
y=f(−1)=(−1)2+(−1)y=f(-1)=(-1)^{2}+(-1)

STEP 6

Calculate the y-value when x=−1x=-1.
y=f(−1)=1−1=0y=f(-1)=1-1=0

STEP 7

Now that we have the points (−3,6)(-3,6) and (−1,0)(-1,0), we can find the slope of the secant line. The slope of a line through two points (x1,y1)(x1,y1) and (x2,y2)(x2,y2) is given by the formulam=y2−y1x2−x1m=\frac{y2-y1}{x2-x1}

STEP 8

Substitute the values of the points into the slope formula.
m=0−6−1−(−3)m=\frac{0-6}{-1-(-3)}

STEP 9

Calculate the slope of the secant line.
m=−62=−3m=\frac{-6}{2}=-3

STEP 10

Now that we have the slope of the secant line, we can find its equation. The equation of a line is given by the formulay−y=m(x−x)y-y=m(x-x)

STEP 11

Substitute the slope and one of the points into the line equation.
y−6=−3(x+3)y-6=-3(x+3)

STEP 12

implify the equation of the secant line.
y=−x−9+6y=-x-9+6y=−x−y=-x-

STEP 13

Now, we need to find the equation of the tangent line when x=−3x=-3. The slope of the tangent line is given by the derivative of the function at that point.
f′(x)=2x+f'(x)=2x+

STEP 14

Substitute x=−3x=-3 into the derivative to find the slope of the tangent line.
f′(−3)=2(−3)+f'(-3)=2(-3)+

STEP 15

Calculate the slope of the tangent line.
f′(−3)=−+=−5f'(-3)=-+=-5

STEP 16

Now that we have the slope of the tangent line, we can find its equation. Substitute the slope and the point (−3,6)(-3,6) into the line equation.
y−6=−5(x+3)y-6=-5(x+3)

STEP 17

implify the equation of the tangent line.
y=−5x−15+6y=-5x-15+6y=−5x−9y=-5x-9a. The equation of the secant line is y=−3x−3y=-3x-3 b. The equation of the tangent line is y=−5x−9y=-5x-9

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