If f(2)=3 and f'(2)=5, find an equation of
a. The tangent line
b. The normal line to the graph of y=f(x) at the point where x=2
So I plotted (2,3) and (2,5)...but that's not getting anywhere...
Thanks
that's the slope because the derivative is stated to be 5 and derivative is slope
y = m x + b
y = 5 x+b
3 = 5*2 + b
b = 3-10 = -7
so
y = 5 x - 7
quick question:
how do you know 5 is the slope?
Well, don't worry, I'm here to help you out and bring some humor into math! Let's tackle this problem with a smile:
a. The tangent line:
To find the equation of the tangent line to the graph of y=f(x) at x=2, we need two things: the slope and a point on the line. The slope of the tangent line is given by f'(2) = 5.
Now, we already have a point on the line, which is (2, 3), so we can use the point-slope formula. Putting it all together, we have:
(y - 3) = 5(x - 2)
There you have it! The equation of the tangent line is y - 3 = 5x - 10. Feel free to simplify it further if you'd like.
b. The normal line:
To find the equation of the normal line, we need to find the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is 5, the slope of the normal line is -1/5.
Now, we can use the point-slope formula again to find the equation of the normal line. Using the point (2, 3) as before, we have:
(y - 3) = -1/5(x - 2)
That's it! The equation of the normal line is y - 3 = -1/5(x - 2). And just like the last equation, feel free to simplify it further if it tickles your funny bone.
Keep up the great work, and always remember to approach math with a smile!
To find the equation of the tangent line to the graph of y = f(x) at the point where x = 2, we need to use the point-slope form of a line. The point-slope form of a line is given by:
y - y1 = m(x - x1)
where (x1, y1) represents a point on the line and m represents the slope of the line. In this case, the point is (2, f(2)) = (2, 3) and the slope is f'(2) = 5.
a. Tangent line: To find the equation of the tangent line, we can plug these values into the point-slope form:
y - 3 = 5(x - 2)
Simplifying, we have:
y - 3 = 5x - 10
Now, let's isolate y:
y = 5x - 7
Therefore, the equation of the tangent line to the graph of y = f(x) at the point where x = 2 is y = 5x - 7.
b. Normal line: The normal line to a graph is perpendicular to the tangent line at a given point. In this case, the slope of the normal line will be the negative reciprocal of the slope of the tangent line.
The slope of the tangent line is 5, so the slope of the normal line is -1/5.
Using the same point (2, 3), we can find the equation of the normal line using the point-slope form:
y - 3 = (-1/5)(x - 2)
Multiplying through by 5, we have:
5y - 15 = -x + 2
Rearranging, we get:
x + 5y = 17
Hence, the equation of the normal line to the graph of y = f(x) at the point where x = 2 is x + 5y = 17.