At what points on the graph of $y={x}^{2}$ does the tangent line pass through (3, -7)?

Jace Baxter
2022-02-11
Answered

At what points on the graph of $y={x}^{2}$ does the tangent line pass through (3, -7)?

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Jaden Petersen

Answered 2022-02-12
Author has **15** answers

The points are (-1, 1) and (7, 49).

Method: Find the general form of the equation of a line tangent to the graph of

$y={x}^{2}$ . Then find the particular points that satisfy the condition: the tangent line passes through (3, -7).

I will continue to use x and y as variables.

Consider a particular value of x, calls it a.

The point an the graph at x=a has coordinates$(a,{a}^{2})$ (The y-coordinate must satisfy $y={x}^{2}$ in order to be on the graph.)

The slope of the tangent to the graph is determined by differentiating: y'=2x, so at the point (a,$a}^{2$ ) the slope of the tangent is m=2a.

Use your favorite tehnique to find the equation of the line through (a,$a}^{2$ ) with slope 2a.

The tangent line has equation:$y=2ax-{a}^{2}$ .

(One way to find the line: start with$y-{a}^{2}=2a(x-a)$ , so $y-{a}^{2}=2ax-2{a}^{2}$ .

Add$a}^{2$ to both sides to get $y=2ax-{a}^{2}$ .)

We have been asked to make the point (3, -7) lie on the line. So we need,

$(-7)=2a\left(3\right)-{a}^{2}$ . Now, solve for a

$-7=6a-{a}^{2}$ if and only if ${a}^{2}-6x-7=0$ .

Solve by factoring: (a+1)(a-7)=0, which requires a=-1 or a=7.

The points we are looking for, then, are (-1, 1) and (7, 49).

You can now check the answers by verifying that the point (3, -7) lies on the lines:

y=-2x-1 (the tangent when a=-1),

and also on y=14x-49 (the tangent when a=7).

Method: Find the general form of the equation of a line tangent to the graph of

I will continue to use x and y as variables.

Consider a particular value of x, calls it a.

The point an the graph at x=a has coordinates

The slope of the tangent to the graph is determined by differentiating: y'=2x, so at the point (a,

Use your favorite tehnique to find the equation of the line through (a,

The tangent line has equation:

(One way to find the line: start with

Add

We have been asked to make the point (3, -7) lie on the line. So we need,

Solve by factoring: (a+1)(a-7)=0, which requires a=-1 or a=7.

The points we are looking for, then, are (-1, 1) and (7, 49).

You can now check the answers by verifying that the point (3, -7) lies on the lines:

y=-2x-1 (the tangent when a=-1),

and also on y=14x-49 (the tangent when a=7).

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