consider the function f(x)=2x^2 +1. its tangent line at x=2 goes through the points (2,y1) and (-1,y2) where y1=?, y2=?
To find the points (2, y1) and (-1, y2) on the tangent line to the function f(x) = 2x^2 + 1 at x = 2, we need to first find the slope of the tangent line at that point.
The slope of the tangent line can be found by taking the derivative of the function f(x) with respect to x and evaluating it at x = 2. Let's do that:
f'(x) = d/dx (2x^2 + 1)
= 4x
Now, let's evaluate f'(x) at x = 2:
f'(2) = 4(2)
= 8
So, the slope of the tangent line at x = 2 is 8.
Using the point-slope form of a linear equation, we can determine the equation of the tangent line:
y - y1 = m(x - x1)
where m is the slope of the tangent line and (x1, y1) are the coordinates of the given point (2, y1).
Plugging in the values, we get:
y - y1 = 8(x - 2)
Simplifying this equation, we have:
y - y1 = 8x - 16
Now, we can substitute the x-coordinate of the other given point (-1, y2) into this equation to find y2.
y2 - y1 = 8(-1) - 16
Simplifying further, we have:
y2 - y1 = -8 - 16
y2 - y1 = -24
Since we know the x-coordinate of the second point (-1), we can substitute it into the equation of the tangent line:
-1 - y1 = 8(-1) - 16
-1 - y1 = -8 - 16
-1 - y1 = -24
Finally, we can solve for y1 and y2:
y1 = -24 + 1
= -23
y2 = -1 - (-24)
= -1 + 24
= 23
Therefore, the points (2, y1) and (-1, y2) on the tangent line are (2, -23) and (-1, 23), respectively.