Find an equation of the straight line having slope 1/4 that is...

Question

Find an equation of the straight line having slope 1/4 that is tangent to the curve y = √x.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says what is the equation of the tangent line to Y is equal to 3x2d at the point where its slope is 3. So this problem wants us to find the equation of the tangent line 2 Y is equal to 3x2 when the slope of that tangent line is equal to 3. So, the first thing we need to do is determine the slope of our tangent line. 2, our equation Y is equal to 3X2d at any point on that curve. So recall that the slope of our tangent line, which we'll label as M tangent, at a point on a curve, is equal to the limit, as H approaches 0 of the quantity of F of X plus H. Minus F of X. In quantity, divided by H. Now, for F of X here, we're going to use 3 X squad. So that means that this is going to be equal to the limit, as H approaches 0 of the quantity of 3, multiplying the quantity of X plus H in quantity squared. Minus 3X2. In quantity, divided by H. So we can multiply out the binomial in the numerator, so this is going to be equal to the limit. As H approaches 0. Of the quantity of 3, multiplying the quantity of X2 plus 2 H X. Plus h 2 in quantity minus 3 x 2 in quantity divided by H Now, when we distribute the 3 into the terms in the parentheses in the numerator, we'll see that the 3 X 2 terms will cancel out. So, what's going to be left is the limit as H approaches 0 of the quantity of 6 H X. Plus 38 squad in the numerator, and that is still going to be divided by H in the denominator. So, we can factor out and cancel the H in the denominator. So we can factor out H out of both terms in the numerator, cancel out the H in the denominator. And so this is going to be equal to the limit, as H approaches 0 of 6 X plus 3 H. And so evaluating the limit by direct substitution, this is going to be equal to 6 X plus 3 multiplied by 0. Which is equal to 6 X. So, we want to determine the point at which the slope of our tangent line is equal to 3. That means that we're going to set 6 X. Equal to 3. And we're going to solve this equation for X. So dividing both sides by 6, we'll get X is equal to 1 divided by 2. Now we need the point, so we need to define the Y value, so we'll just substitute into our expression for Y. So we'll have Y is equal to 3, multiplying the quantity of 1 divided by 2 in quantity squared. And evaluating this expression, this is going to be equal to. 3 divided by 4. So, we actually have a point on a curve where the tangent line at this point has a slope of 3, and that point is 1 divided by 23 divided by 4. So now we need to do is find the equation for the tangent line at this point, and here we'll just use the point slope form of a line. So, we'll have Y minus 3 divided by 4. is equal to our slope, which is 3, multiplying the quantity of X minus 1 divided by 2 in quantity. So distributing the 3 on the right-hand side, we will have Y minus 3 divided by 4 is equal to 3X minus 3 divided by 2, and then solving for Y, we'll just add 3 divided by 4 to both sides of the equation, and we will get Y is equal to 3 X minus 3 divided by 4. And so this is the equation for the tangent line that we were looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Find an equation of the straight line having slope 1/4 that is tangent to the curve y = √x.

Key Concepts

Tangent Line

Derivative

Point-Slope Form

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