The slope of the tangent line to the parabola y=4x2–6x+6 at the point where x=5 is:
The equation of this tangent line can be written in the form y=mx+b where m is:
and where b is:
Well, let's find the slope of the tangent line to the parabola at the point where x = 5. To do that, we need to find the derivative of the parabola:
y = 4x^2 - 6x + 6
Take the derivative of y with respect to x:
dy/dx = 8x - 6
Now we can substitute x = 5 into the derivative to find the slope:
m = 8(5) - 6
m = 40 - 6
m = 34
So, the slope of the tangent line is 34. Now, let's find the equation of the tangent line in the form y = mx + b, where m is the slope and b is the y-intercept. We already have the value of m, so we just need to find b.
To find b, we can substitute the point (5, y) into the equation and solve it for b:
y = mx + b
6 = 34(5) + b
6 = 170 + b
b = 6 - 170
b = -164
Therefore, the equation of the tangent line is y = 34x - 164, where m = 34 and b = -164.
To find the slope of the tangent line to the parabola at the point where x = 5, we need to find the derivative of the function y = 4x^2 - 6x + 6.
Differentiating the function with respect to x, we get:
dy/dx = 8x - 6
Now plug x = 5 into the derivative to find the slope at x = 5:
dy/dx = 8(5) - 6
= 40 - 6
= 34
Therefore, the slope of the tangent line at the point where x = 5 is 34.
To find the equation of the tangent line in the form y = mx + b, we can use the point-slope form of a line and substitute the coordinates of the point where x = 5 into the equation.
Using the point (5, y) where x = 5, we substitute x = 5 and y = 4(5)^2 - 6(5) + 6 = 64 into the equation y = mx + b:
64 = 34(5) + b
Simplifying this equation, we get:
64 = 170 + b
Rearranging the equation to solve for b, we have:
b = 64 - 170
b = -106
Therefore, the equation of the tangent line in the form y = mx + b, where m is the slope (34) and b is the y-intercept (-106), is:
y = 34x - 106.
To find the slope of the tangent line to the parabola at the point where x=5, you need to differentiate the given parabola equation with respect to x and then evaluate it at x=5.
First, let's find the derivative of the parabola equation y=4x^2-6x+6.
The derivative of each term can be found using the power rule:
d/dx (x^n) = n*x^(n-1)
For y=4x^2-6x+6, taking the derivative of each term:
dy/dx = d/dx (4x^2) - d/dx (6x) + d/dx (6)
dy/dx = 8x^(2-1) - 6*1^(1-1) + 0
Simplifying further:
dy/dx = 8x - 6
Now, to find the slope at the point where x=5, substitute x=5 into the derivative equation:
m = dy/dx (x=5) = 8*(5) - 6 = 40 - 6 = 34
The slope of the tangent line is 34.
To find the equation of the tangent line in the form y=mx+b, we also need to find the y-intercept (b).
To find b, we use the point-slope form of a line y-y1 = m(x-x1), where (x1, y1) is the point of tangency (x=5, y=?).
Substituting x=5 and y=? into the equation:
y - y1 = m(x - x1)
y - y1 = 34(x - 5)
Since the point of tangency is x=5, we have:
y - y1 = 34(x - 5)
y - y1 = 34x - 170
To simplify, let's move y1 and rewrite it as b:
y = 34x - 170 + y1
So, the equation of the tangent line in the form y=mx+b is:
y = 34x + (y1 - 170)
Therefore, the value of m (the slope) is 34 and the value of b is (y1 - 170)
The slope of the tangent at that point will be the same as the slope of the parabola, by definition.
How do you get the slope of the parabola at that point? Differentiate it. What is the differential of 4x^2–6x+6? Now plug in x=5 to your answer, and the arithmetic gives you the slope.
y = mx + b. m is your slope you just got. We need a point on the line to get b.
Where x=5, the point where the line touches the parabola, y=76, so we have the point (5,76).
76=170+b
and that gives you b.