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 of the function f(x) = 2x^2 + 1 at x = 2, we need to find the corresponding y-values.
We can start by finding the slope of the tangent line at x = 2, which is equal to the derivative of the function evaluated at that point.
1. Differentiate the function f(x) = 2x^2 + 1 with respect to x:
f'(x) = 4x
2. Substitute x = 2 into the derivative:
f'(2) = 4(2) = 8
So, the slope of the tangent line at x = 2 is 8.
Since we know a point on the tangent line (2, y1), we can use the point-slope form of a line to find the equation of the tangent line and then determine y1.
3. Use the point-slope form of a line:
y - y1 = m(x - x1)
y - y1 = 8(x - 2)
4. Substitute the x-coordinate and y-coordinate of the known point (2, y1) into the equation above:
y - y1 = 8(x - 2)
y - y1 = 8x - 16
5. Simplify the equation:
y = 8x - 16 + y1
Now we have the equation of the tangent line at x = 2.
To find y1, we plug x = 2 into the original function f(x) = 2x^2 + 1:
f(2) = 2(2)^2 + 1
f(2) = 2(4) + 1
f(2) = 8 + 1
f(2) = 9
Therefore, y1 = 9.
To find y2, we need to find the y-value of the tangent line at x = -1.
6. Substitute x = -1 into the equation of the tangent line:
y = 8x - 16 + y1
y = 8(-1) - 16 + 9
y = -8 - 16 + 9
y = -15
Therefore, y2 = -15.
In conclusion, the points on the tangent line of the function f(x) = 2x^2 + 1 at x = 2 are (2, 9) and (-1, -15).