Let f(x) = 3x^2 - 2x + 1
a) write an equation of the line going through the points (1, f(1)), and (2, f(2)).
b) Find a point on the graph of f where the tangent line to the graph has the same slope as the line in part (a). Write the equation of the tangent line at that point.
I got the answer to part A. For part b, I'm confused...isn't the slope of the tangent line -1/slope of normal line?
for part a) i got y = 7x -5
which means you evaluated the slope as 7. So where else is it seven?
f'=6x-2=7
x=1.5
y=3(9/6)^2-2(9/6)+1 figure that out.
Now , you know x,y and the slope, so figure the intercept b.
y=mx+b
Thank you!
To write the equation of the line going through the points (1, f(1)) and (2, f(2)), you can use the point-slope form of a linear equation.
a)
Let's start by finding the y-coordinates of the points (1, f(1)) and (2, f(2)):
For the point (1, f(1)):
Substitute x = 1 into the equation f(x) = 3x^2 - 2x + 1:
f(1) = 3(1)^2 - 2(1) + 1
f(1) = 3 - 2 + 1
f(1) = 2
So, the y-coordinate of the point (1, f(1)) is 2.
For the point (2, f(2)):
Substitute x = 2 into the equation f(x) = 3x^2 - 2x + 1:
f(2) = 3(2)^2 - 2(2) + 1
f(2) = 12 - 4 + 1
f(2) = 9
So, the y-coordinate of the point (2, f(2)) is 9.
Now that we have the coordinates of two points on the line, we can use the point-slope form:
y - y1 = m(x - x1)
where (x1, y1) is one of the given points, and m is the slope.
Let's use the point (1, 2):
Using the slope formula, we can find the slope between the two points:
m = (f(2) - f(1))/(2 - 1)
= (9 - 2)/(2 - 1)
= 7/1
= 7
So, the slope of the line passing through the points (1, f(1)) and (2, f(2)) is 7.
Now we can substitute the values into the point-slope form equation:
y - 2 = 7(x - 1)
Simplifying:
y - 2 = 7x - 7
Rearranging to slope-intercept form (y = mx + b):
y = 7x - 7 + 2
y = 7x - 5
Therefore, the equation of the line going through the points (1, f(1)) and (2, f(2)) is y = 7x - 5.
b)
To find a point on the graph of f(x) where the tangent line has the same slope as the line in part (a), we need to find the value of x that satisfies that condition.
Recall that the slope of the tangent line to the graph at a specific point is given by the derivative of the function at that point. Therefore, we need to find the derivative of f(x) and solve for a point where the derivative is equal to 7.
First, let's find the derivative of f(x):
f'(x) = 6x - 2
Now, to find the x-coordinate of the point where the tangent line has the same slope as the line in part (a), we set the derivative equal to 7 and solve for x:
6x - 2 = 7
Adding 2 to both sides:
6x = 9
Dividing both sides by 6:
x = 9/6
Simplifying:
x = 3/2
So, the x-coordinate of the point where the tangent line has the same slope as the line in part (a) is 3/2.
To find the corresponding y-coordinate, substitute x = 3/2 into the original function:
f(3/2) = 3(3/2)^2 - 2(3/2) + 1
f(3/2) = 3(9/4) - 3 + 1
f(3/2) = 27/4 - 3 + 1
f(3/2) = (27 - 12 + 4)/4
f(3/2) = 19/4
Therefore, the point on the graph of f(x) where the tangent line has the same slope as the line in part (a) is (3/2, 19/4).
To write the equation of the tangent line at that point, we can use the point-slope form again. The slope of the tangent line is equal to the derivative at that point, which is 6(3/2) - 2 = 7.
Using the point (3/2, 19/4), we have:
y - (19/4) = 7(x - 3/2)
Simplifying:
y - (19/4) = 7x - 21/2
y = 7x - 21/2 + 19/4
y = 7x - 21/2 + 38/4
y = 7x - 21/2 + 38/4
y = 7x - 21/2 + 19/4
y = 7x - 26/4
Finally, simplifying further:
y = 7x - 13/2
This is the equation of the tangent line to the graph of f(x) at the point (3/2, 19/4), where the slope of the tangent line is the same as the slope of the line in part (a).