Consider the function f(x)=7x^(2)+9.
Its tangent line at x=5 goes through the points (1,y1) and (−7,y2) where y1= ___ , and y2= ___
y = 7x^2 + 9
dy/dx = 14x which is the slope of the tangent
so when x = 5, slope = 70
and y = 7(25) + 9 = 184
so the tangent has a slope of 70 and goes through the point (5,184)
equation of tangent:
y - 184 = 70(x-5)
y = 70x - 166
to get y1, replace x with 1
to get y2, replace x with -7
Thank you very much! I was stuck and needed this.
To find the y-values (y1 and y2) of the points where the tangent line at x=5 goes through, we need to find the slope of the tangent line and use the point-slope form of a line to determine the y-intercept.
First, let's find the slope of the tangent line by taking the derivative of the function f(x) with respect to x:
f'(x) = d/dx (7x^2 + 9)
To differentiate 7x^2, we apply the power rule, which states that the derivative of x^n is n*x^(n-1).
f'(x) = 7*2x^(2-1) + 0
= 14x
Evaluating the derivative at x=5:
f'(5) = 14*5
= 70
Therefore, the slope of the tangent line at x=5 is 70.
Next, we can use the point-slope form of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.
For the first point (1, y1), we have:
y - y1 = 70(x - 5)
Expanding:
y - y1 = 70x - 350
Since this line passes through (1, y1), we can substitute the coordinates (1, y1) and solve for y1:
y1 - y1 = 70(1) - 350
0 = 70 - 350
70 = 350
Therefore, y1 = 420.
Now, for the second point (-7, y2):
y - y2 = 70(x - 5)
Expanding:
y - y2 = 70x - 350
Since this line passes through (-7, y2), we can substitute the coordinates (-7, y2) and solve for y2:
y2 - y2 = 70(-7) - 350
0 = -490 - 350
840 = -840
Therefore, y2 = -840.
The values of y1 and y2 are:
y1 = 420 and y2 = -840.
To find the points (1, y1) and (-7, y2) on the tangent line to the function f(x) at x = 5, we need to determine the coordinates of these points.
To begin, let's find the derivative of the function f(x) = 7x^2 + 9, which will give us the slope of the tangent line at any given point:
f'(x) = d/dx (7x^2 + 9)
= 14x
Next, let's plug in x = 5 into f'(x) to find the slope of the tangent line at x = 5:
f'(5) = 14(5)
= 70
Now that we have the slope of the tangent line, we can use the point-slope form of a linear equation to find the equation of the tangent line, which passes through the point (5, f(5)) = (5, 7(5)^2 + 9):
y - y1 = m(x - x1)
where m represents the slope of the tangent line, and (x1, y1) represents the point where the tangent line touches the function.
Plugging in the values, we have:
y - (7(5)^2 + 9) = 70(x - 5)
Simplifying:
y - (7(25) + 9) = 70(x - 5)
y - (175 + 9) = 70(x - 5)
y - 184 = 70x - 350
y = 70x - 350 + 184
y = 70x - 166
Now we can find the y-coordinates of the points (1, y1) and (-7, y2) on the tangent line by plugging in x = 1 and x = -7:
For (1, y1):
y1 = 70(1) - 166
Simplifying:
y1 = 70 - 166
y1 = -96
Therefore, y1 = -96.
For (-7, y2):
y2 = 70(-7) - 166
Simplifying:
y2 = -490 - 166
y2 = -656
Therefore, y2 = -656.
In summary, the coordinates of the points on the tangent line to the function f(x) at x = 5 are (1, -96) and (-7, -656).