Question

59 - 62. equations of tangent lines
a. find an equation of the line tangent to the given curve at a.
b. use a graphing utility to graph the curve and the tangent line on the same set of axes.

1. y = - 3x^2 + 2; a = 1
2. y = x^3 - 4x^2 + 2x - 1; a = 2
3. y = e^x; a = ln 3
4. y = \frac{e^x}{4}-x; a = 0

Answer:

1. For problem 59 ($y = - 3x^{2}+2$, $a = 1$):

• Step - by - Step Format
• Step1: Find the derivative of the function
• Given $y=-3x^{2}+2$, using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $y^\prime=\frac{d}{dx}(-3x^{2}+2)=-6x$.
• Step2: Find the slope of the tangent line at $x = a$
• Substitute $x = 1$ into the derivative. When $x = 1$, $y^\prime(1)=-6(1)=-6$. So the slope $m=-6$.
• Step3: Find the point on the curve at $x = a$
• Substitute $x = 1$ into the original function $y=-3(1)^{2}+2=-3 + 2=-1$. So the point $(x_0,y_0)=(1,-1)$.
• Step4: Use the point - slope form of a line to find the equation of the tangent line
• The point - slope form is $y - y_0=m(x - x_0)$. Substituting $m=-6$, $x_0 = 1$, and $y_0=-1$ gives $y+1=-6(x - 1)$.
• Expand to get $y+1=-6x + 6$, or $y=-6x+5$.
• Answer: The equation of the tangent line is $y=-6x + 5$.

1. For problem 60 ($y=x^{3}-4x^{2}+2x - 1$, $a = 2$):

• Step1: Find the derivative of the function
• Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, $y^\prime=\frac{d}{dx}(x^{3}-4x^{2}+2x - 1)=3x^{2}-8x + 2$.
• Step2: Find the slope of the tangent line at $x = a$
• Substitute $x = 2$ into the derivative. $y^\prime(2)=3(2)^{2}-8(2)+2=12-16 + 2=-2$. So the slope $m=-2$.
• Step3: Find the point on the curve at $x = a$
• Substitute $x = 2$ into the original function $y=(2)^{3}-4(2)^{2}+2(2)-1=8-16 + 4-1=-5$. So the point $(x_0,y_0)=(2,-5)$.
• Step4: Use the point - slope form of a line to find the equation of the tangent line
• Using $y - y_0=m(x - x_0)$ with $m=-2$, $x_0 = 2$, and $y_0=-5$, we get $y + 5=-2(x - 2)$.
• Expand to get $y+5=-2x + 4$, or $y=-2x-1$.
• Answer: The equation of the tangent line is $y=-2x-1$.

1. For problem 61 ($y = e^{x}$, $a=\ln3$):

• Step1: Find the derivative of the function
• The derivative of $y = e^{x}$ is $y^\prime=e^{x}$.
• Step2: Find the slope of the tangent line at $x = a$
• Substitute $x=\ln3$ into the derivative. $y^\prime(\ln3)=e^{\ln3}=3$. So the slope $m = 3$.
• Step3: Find the point on the curve at $x = a$
• Substitute $x=\ln3$ into the original function $y = e^{\ln3}=3$. So the point $(x_0,y_0)=(\ln3,3)$.
• Step4: Use the point - slope form of a line to find the equation of the tangent line
• Using $y - y_0=m(x - x_0)$ with $m = 3$, $x_0=\ln3$, and $y_0 = 3$, we get $y-3=3(x-\ln3)$.
• Expand to get $y=3x-3\ln3 + 3$.
• Answer: The equation of the tangent line is $y=3x-3\ln3 + 3$.

1. For problem 62 ($y=\frac{e^{x}}{4}-x$, $a = 0$):

• Step1: Find the derivative of the function
• Using the sum - rule and the derivative of $e^{x}$, $y^\prime=\frac{1}{4}e^{x}-1$.
• Step2: Find the slope of the tangent line at $x = a$
• Substitute $x = 0$ into the derivative. $y^\prime(0)=\frac{1}{4}e^{0}-1=\frac{1}{4}-1=-\frac{3}{4}$. So the slope $m=-\frac{3}{4}$.
• Step3: Find the point on the curve at $x = a$
• Substitute $x = 0$ into the original function $y=\frac{e^{0}}{4}-0=\frac{1}{4}$. So the point $(x_0,y_0)=(0,\frac{1}{4})$.
• Step4: Use the point - slope form of a line to find the equation of the tangent line
• Using $y - y_0=m(x - x_0)$ with $m=-\frac{3}{4}$, $x_0 = 0$, and $y_0=\frac{1}{4}$, we get $y-\frac{1}{4}=-\frac{3}{4}(x - 0)$.
• Simplify to get $y=-\frac{3}{4}x+\frac{1}{4}$.
• Answer: The equation of the tangent line is $y=-\frac{3}{4}x+\frac{1}{4}$.