Find the equation of the tangent to the graph at the indicated point.
f(x) = x^2 − 1; a = 3
and
f(x) = x^2 − 8x; a = −9
whats the difference between 1 and 8x. What formula do I use and how should i solve both of these problems?
Thank you
Why are you using ""a , as in a = 3 ?
There is no "a" in either equation
Is this a "first principle" question to find the equation of the tangent to a curve ?
I will assume that is the case, and do the 2nd problem, the harder of the two .....
slope of tangent at x = -9 :
f(-9) = 81 -8(-9) = 153
f(-9 +h) = (-9+h)^2 -8(-9+h)
= 81 - 18h + h^2 +72 - 8h
= h^2 -26h + 153
slope
=lim [ f(-9+h) - f(-9) ]/h , as h ---> 0
= limit [ (h^2 - 26h + 153) - 153]/h , as h --> 0
= limit [ h(h - 26 ]/h
= limi h-26 , as h --> 0
= -26
To find the equation of the tangent to the graph, you need to use the formula for the slope of a tangent line at a given point. The slope of the tangent line is equal to the derivative of the function evaluated at the given point. Once you have the slope, you can use the point-slope form of a line to find the equation of the tangent.
For the first problem, let's find the equation of the tangent to the function f(x) = x^2 − 1 at the point where x = 3.
Step 1: Find the derivative of the function f(x) with respect to x.
The derivative of f(x) = x^2 − 1 would be 2x.
Step 2: Evaluate the derivative at the point x = 3.
Substitute x = 3 into the derivative 2x:
2 * 3 = 6
So, the slope of the tangent line at x = 3 is 6.
Step 3: Use the point-slope form of a line to find the equation of the tangent.
The equation of a line in point-slope form is given by: y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.
Using the point where x = 3, we have (3, f(3)) = (3, 3^2 - 1) = (3, 8), and the slope m = 6.
Substituting these values into the point-slope form, we get:
y - 8 = 6(x - 3)
Simplifying:
y - 8 = 6x - 18
y = 6x - 10
So, the equation of the tangent to the graph of f(x) = x^2 − 1 at the point where x = 3 is y = 6x - 10.
Now, let's solve the second problem:
f(x) = x^2 − 8x; a = -9
Step 1: Find the derivative of the function f(x) with respect to x.
The derivative of f(x) = x^2 − 8x would be 2x - 8.
Step 2: Evaluate the derivative at the point x = -9.
Substitute x = -9 into the derivative 2x - 8:
2 * (-9) - 8
-18 - 8
-26
So, the slope of the tangent line at x = -9 is -26.
Step 3: Use the point-slope form of a line to find the equation of the tangent.
The point given is (-9, f(-9)) = (-9, (-9)^2 - 8(-9)) = (-9, 81 + 72) = (-9, 153), and the slope is -26.
Substituting these values into the point-slope form, we have:
y - 153 = -26(x - (-9))
y - 153 = -26(x + 9)
Simplifying:
y - 153 = -26x - 234
y = -26x - 81
So, the equation of the tangent to the graph of f(x) = x^2 − 8x at the point where x = -9 is y = -26x - 81.
To summarize, for both problems:
1. Find the derivative of the function with respect to x.
2. Evaluate the derivative at the given point to find the slope of the tangent line.
3. Use the point-slope form of a line to find the equation of the tangent, substituting the given point and the slope obtained in step 2.