What is the equation of the line tangent to the function f(x)=-x^2+5 at the point (1,4)?
A. y=-x+5
B. y=-x+9
C. y=-2x+6
D y=-2x+9
Noooo!
oobleck just told you the slope is -2, so ......
use your y-y1=m(x-x1)
simplify it, and change it to the form y = mx + b
From that, I got C.
Rate of change quick check
1. D
2. A
3. C
4. C
the slope is -2x = -2
so now use the point-slope form to choose C or D
@ oobleck so that means that the slope is 1. And if point-slope form is y-y1=m(x-x1), wouldn't that give me y-4=x-1?
Your answers are correct! Good job!
To find the equation of the line tangent to a function at a given point, you need to find the derivative of the function at that point.
In this case, we have the function f(x) = -x^2 + 5, and we want to find the tangent line at the point (1,4).
To find the derivative of f(x), we differentiate the function with respect to x. The derivative of -x^2 is -2x, and the derivative of 5 is 0 (as constants have a derivative of 0). Therefore, the derivative of f(x) is f'(x) = -2x.
Next, we evaluate the derivative at x = 1 to find the slope of the tangent line at that point. Substituting x = 1 into f'(x), we get f'(1) = -2(1) = -2.
So, the slope of the tangent line at the point (1,4) is -2.
Now, we can use the point-slope form of a linear equation, which states that y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.
Using the point (1,4) and the slope -2, we have y - 4 = -2(x - 1).
Simplifying this equation, we get y - 4 = -2x + 2.
Finally, we can rearrange this equation to the standard form y = mx + b, where m is the slope and b is the y-intercept. By adding 4 to both sides of the equation, we obtain y = -2x + 6.
Comparing this equation with the answer choices, we can see that the correct equation is:
C. y = -2x + 6.