Find an equation for the tangent line of y = e 2 x that is parallel to the linear equation y = 7 x + 13 . Provide your answer in slope–inter

Question

Find an equation for the tangent line of y = e 2 x that is parallel to the linear equation y = 7 x + 13 . Provide your answer in slope–intercept form of a linear equation, y = m x + b , where m is the slope and b is the y – intercept. Express m and b as exact numbers.

in progress 0
Hadley 23 hours 2021-09-15T00:28:53+00:00 1 Answer 0

Answers ( )

    0
    2021-09-15T00:30:52+00:00

    Answer:

    y=2e^{2x}(x-\frac{\text{ln}(\frac{7}{2})}{2})+3.5

    Step-by-step explanation:

    We are asked to find equation of tangent line of function y=e^{2x} that is parallel to the linear equation y = 7x+13.

    First of all, we will find the derivative of given function applying chain rule as:

    y'=\frac{d}{dx}(e^{2x})\times \frac{d}{dx}(2x)

    y'=e^{2x}\times(2)

    y'=2e^{2x}

    Since derivative represents slope or rate of change, so we will equate derivative to slope of line  y = 7x+13 and solve for x as:

    2e^{2x}=7

    e^{2x}=\frac{7}{2}

    \text{ln}(e^{2x})=\text{ln}(\frac{7}{2})

    2x\text{ln}(e)=\text{ln}(\frac{7}{2})

    2x(1)=\text{ln}(\frac{7}{2})

    x=\frac{\text{ln}(\frac{7}{2})}{2}

    Now, we will substitute x=\frac{\text{ln}(\frac{7}{2})}{2} in equation y=e^{2x} to find value of y as:

    y=e^{2(\frac{\text{ln}(\frac{7}{2})}{2})}

    y=e^{\text{ln}(\frac{7}{2})}

    Applying log property a^{\text{log}_a(b)}=b, we will get:

    y=\frac{7}{2}=3.5

    Now, we will substitute m=2e^{2x}, x=\frac{\text{ln}(\frac{7}{2})}{2} and y=3.5 in point slope form as:

    y-3.5=2e^{2x}(x-\frac{\text{ln}(\frac{7}{2})}{2})

    y=2e^{2x}(x-\frac{\text{ln}(\frac{7}{2})}{2})+3.5

    Therefore, the equation of the tangent line would be y=2e^{2x}(x-\frac{\text{ln}(\frac{7}{2})}{2})+3.5.

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )