find-equations-of-the-tangent-line-and-normal-line-to-the-curve-at-the-given-point-y-x4-2ex-02

Question

Find equations of the tangent line and normal line to the curve at the given point.
$ y = x^4 + 2e^x, (0,2) $

Clarissa Noh · Accepted Answer

it&#x27;s clear. So when you right here. So we have y is equal to X to the fourth plus to eat tax. We&#x27;re gonna differentiate, do you? I over d x and we get d over t X in terms of X for X to the fourth plus to eat the axe. This gives us four X cute plus to eat the axe in the slope of the tangent at zero comma, too. It is going to be, too, because when we plug in zero, this cross is out and this becomes just to so it&#x27;s too. So the slope of the normal line has to be negative 1/2 since the product of slopes of perpendicular lines is negative. One. The tangent is the line. Passing through 02 was slope too. So the equation is lie minus two over X minus zero is equal to two. We get why it&#x27;s equal to two acts plus two, and the normal is the line passing through 02 with slope negative 1/2. So the equation is boy minus two over X minus. Cerro is equal to negative 1/2. Why is equal to negative 1/2 X plus two. So this is our tangent, and this is our normal

Problem

Find equations of the tangent line and normal lin…

Problem 37 Medium Difficulty

Find equations of the tangent line and normal line to the curve at the given point.
$ y = x^4 + 2e^x, (0,2) $

Answer

$$y-2=-\frac{1}{2}(x-0) \text { or } y=-\frac{1}{2} x+2$$

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