please help! Find an equation of the tangent line to the curve at the point (-1, 1).
y = 5x3 - 6x.
y=? ...............i keep getting y=x+2 and its wrong. can someone please help?
Let f(x)=5x³-6x, then
f(-1)=5(-1)³-6(-1)=1
Slope at (x, f(x)) is given by dy/dx=f'(x)
f'(x) = dy/dx = 15x³-6
f'(-1)= 15(-1)²-6=9
Therefore you need a line with a slope of 9 to pass through the point (-1,1).
The standard equation for this is:
(y-y1)=m(x-x1)
where m=9, y1=-1, x1=1
(y-(-1)) = 9(x-1)
y=9x-9-1
y=9x-10 (slope of 9 and passes through (-1,1))
To find the equation of the tangent line to the curve at a given point, you need to use the derivative of the function. Here's how you can find the equation of the tangent line to the curve y = 5x^3 - 6x at the point (-1, 1):
Step 1: Find the derivative of the function.
To find the derivative, you need to differentiate the function with respect to x. Let's differentiate y = 5x^3 - 6x term by term:
dy/dx = d(5x^3)/dx - d(6x)/dx
= 15x^2 - 6
Step 2: Substitute x = -1 into the derivative.
To find the slope of the tangent line at x = -1, substitute x = -1 into the derivative:
m = dy/dx = 15(-1)^2 - 6
= 15 - 6
= 9
Step 3: Use the point-slope form to find the equation of the tangent line.
The equation of a line in point-slope form is given by:
y - y1 = m(x - x1)
where (x1, y1) is the given point (-1, 1) and m is the slope we found in Step 2.
Plugging in the values:
y - 1 = 9(x - (-1))
y - 1 = 9(x + 1)
y - 1 = 9x + 9
Step 4: Simplify the equation.
To get the equation in slope-intercept form (y = mx + b), simplify the equation:
y = 9x + 9 + 1
y = 9x + 10
Therefore, the equation of the tangent line to the curve y = 5x^3 - 6x at the point (-1, 1) is y = 9x + 10.