Find the equation of the tangent to the curve y=1/2x^2-4x-3at the point where the curve intersects the y-axis.
I have now done 2 of your quite straightforward questions. You MUST know how to do these type of relatively easy Calculus problems.
Show me the steps of your solution so I can check it.
I'll help you with the calculus:
y'(x) = x-4
The rest is just Algebra I.
To find the equation of the tangent to the curve at the point where it intersects the y-axis, we need to find the slope of the tangent and the coordinates of the point of tangency.
First, let's find the y-coordinate of the point of tangency. When a curve intersects the y-axis, the x-coordinate is always 0. So, we substitute x = 0 into the equation of the curve to find the y-coordinate:
y = (1/2)(0^2) - 4(0) - 3
y = 0 - 0 - 3
y = -3
Therefore, the point of tangency is (0, -3).
Next, let's find the slope of the tangent at that point. The slope of a curve at any point can be found by taking the derivative of the equation of the curve. The derivative of y with respect to x, denoted as dy/dx, represents the slope of the tangent at any given point.
To find dy/dx, we differentiate the equation of the curve with respect to x. The derivative of x^n is nx^(n-1), and the derivative of a constant is 0. Hence, differentiating the equation y = (1/2)x^2 - 4x - 3 gives us:
dy/dx = 1/2 * 2x - 4
dy/dx = x - 4
Now that we have the slope of the tangent, which is dy/dx = x - 4, and the coordinates of the point of tangency, which is (0, -3), we can use the point-slope form of a linear equation to find the equation of the tangent.
The point-slope form is given by y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Plugging in the values:
y - (-3) = (x - 0)(0 - 4)
y + 3 = -4x
Rearranging the equation to the standard form gives us:
4x + y + 3 = 0
Therefore, the equation of the tangent to the curve at the point where it intersects the y-axis is 4x + y + 3 = 0.