Given the function f and the point Q, find all points P on the...

Question

Given the function f and the point Q, find all points P on the graph of f such that the line tangent to f at P passes through Q. Check your work by graphing f and the tangent lines.

f(x)=x²+1; Q(3, 6)

Accepted Answer

Welcome back, everyone. The function k and point V are given. Determine all points P on the graph of K such that the line tangent tok at P passes through B. KX is X2 + 2 x + 1 and V is a point with X of 4 and. Y 9 were given 4 answer choices. A says -5.16 and 2.9, B 3.4, and 8.81. C 0.1 and 2.9, and D 0.1, 8.81. So first of all, what we're going to do is consider this function and the slope of the tangent line. Now what we're going to do is use the definition. The slope m of the tangent line is equal to the limit, as H approaches 0 of the function k at point X plus H minus the function k at point X, all of that divided by H. So now let's substitute. We're going to get the limit as H approaches 0 of. Now for every X that we have, we're going to replace it with X + H. So we get X + H2 + 2 multiplied by X + H. Plus 1 and then we are subtracting the original function X2 + 2 x + 1. All of that divided by H. Now let's keeps simplifying. We got limit as H approaches 0 of X2 + 2 X H. Plus H2d distributing, we're going to get 2 X2 H, so plus 2 X plus 2 H + 1, then we are subtracting X2 minus 2X minus 1. All of that divided by h we can cancel out a lot of terms. Those are X squared. Then we can cancel out 2 Xs and we can cancel out 1. And now what we're going to do is rewrite the limits. We get limit as H approaches 0. Let's factor out the H, and we're going to get 2 X plus H + 2. All of that divided by H, and this allows us to eliminate the whole H and H in the numerator and the denominator respectively. So now let's attempt the direct substitution. We get 2 x plus H which approaches 0 + 2, and we have got the equation of the slope of the tangent line in a general form which is 2 x + 2. And now we should recall that the slope is equal to the difference in Y divided by the difference in X. So we can take any two points that we want, right? We are going to take two Y coordinates, Y2 minus Y1. And we're going to divide that by Two X coordinates X2 minus X1. What we have to understand is that we're given one of our points, right? So we're given X1, which is 4. This comes from the point B, and Y1, the corresponding Y coordinate is 9. We can take any other point that belongs to the curve K of X, right, and this means that X2 is going to be any X, so we're going to say X and Y2, well, essentially this is KX itself, right? We're just substituting X for every X so nothing really changes. That's X2 + 2 x + 1. And then we're going to use this equation and replace everything that we're given. We already know the slope. 2 X + 2, and this slope must be equal to the difference between Y2 and Y1, so X2 + 2 X + 1 minus. Why 1, which is 9. And this has to be divided by X2 minus X1. So X2 is X. And X1 is 4. Now this is the equation that we want to solve. So let's simplify it. We're going to get X2 + 2 X. 1 minus 9 gives us -8. And we can multiply the denominator by 2 X + 2 to get the right hand side. So we're going to get 2 X. Plus 2 multiplied by x minus 4 under the assumption that x is not equal to 4, right, because that would make the denominator equal to 0, so we can write that x is not equal to 4. And now let's solve it. X2 + 2X minus 8 must be equal to. Now, what we're going to do from here is basically consider our quadratic terms, right? So, Let's apply the foil property to X multiplied by X is 2 X2. Then we're multiplying 2 X by -4. This gives us -8 X. Moving on to 2, we get plus 2X minus 8. And now let's rewrite it in a standard form. First of all, let's subtract X2d from both sides. We're going to get X2. Then we have 8 X plus 2X, which is -6X, and we're also subtracting 2 X, so that's -8X. And -8, well, it can be canceled out because this is the same term on both sides. And this must be equal to 0 because we have subtracted all of the terms from the left hand side. So if we factor it out, we're going to get X multiplied by x minus 8. Equals 0, which gives us two solutions. Either X is equal to 0. Or the second possible value of X is 8. So we want to identify the corresponding Y values. And we can do that by substituting those Xes into K of X. So let's evaluate K at 0. This gives us 02 + 2 multiplied by 0, which is 0. So that's simply 1. And we also want to identify K. At 0.8, so we need to substitute 8. We get 82 + 2 multiplied by 8 + 1, 64 + 165, 65 plus 16. Well, this gives us 81. So we have the required points 0.1 and 8.81. And this corresponds to the answer choice D. Thank you for watching.

Given the function f and the point Q, find all points P on the graph of f such that the line tangent to f at P passes through Q. Check your work by graphing f and the tangent lines.
f(x)=x²+1; Q(3, 6)

Key Concepts

Tangent Line

Finding Points on a Graph

Graphing Functions and Tangents

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