Find c so y=4x+3 is tangent to the curve y=x^2+c
7
Well, finding the value of c in this case requires finding the derivative of the curve y=x^2+c. By taking the derivative of y with respect to x, we get y' = 2x.
Since we want the line y=4x+3 to be tangent to the curve, the slopes of the line and the curve at the point of tangency should be equal. So we set the derivative of the curve equal to the slope of the line:
2x = 4
Solving for x, we get x = 2.
Now, substituting x = 2 back into the original equation of the curve, y = x^2 + c, we can solve for c:
y = 2^2 + c
y = 4 + c
Since the line y = 4x + 3 is tangent to the curve, the value of y at x = 2 must be equal for both equations:
4x + 3 = 4 + c
Plugging in x = 2, we solve for c:
4(2) + 3 = 4 + c
8 + 3 = 4 + c
11 = 4 + c
11 - 4 = c
7 = c
So, the value of c that makes y = 4x + 3 tangent to the curve y = x^2 + c is 7.
And that's how you find the value of c! However, I must say, it's quite a tangent-al process!
To find the value of c such that the line y = 4x + 3 is tangent to the curve y = x^2 + c, we need to determine the points at which the line and the curve intersect.
First, let's set the line and the curve equal to each other and solve for x:
4x + 3 = x^2 + c
Rearranging the equation:
x^2 - 4x + (c - 3) = 0
For the line to be tangent to the curve, there should be only one solution to this quadratic equation. This means that its discriminant should be zero.
The discriminant (D) of a quadratic equation ax^2 + bx + c = 0 is given by D = b^2 - 4ac.
In our equation, the coefficients are: a = 1, b = -4, and c = (c - 3).
Setting the discriminant to zero:
D = (-4)^2 - 4(1)(c - 3) = 0
16 - 4(c - 3) = 0
16 - 4c + 12 = 0
-4c + 28 = 0
-4c = -28
c = 7
Therefore, when c = 7, the line y = 4x + 3 is tangent to the curve y = x^2 + 7.
2
4
you need two things:
x^2+c = 4x+3 (curve intersects line)
2x = 4 (slopes match up)
Now it's easy to find c, right?