The slope of the tangent line to the parabola y=4x2–3x+5 at the point where x=–5 is: (-43)
The equation of this tangent line can be written in the form y=mx+b where m is: (-43)
and where b is:_____________?
To find the equation of the tangent line to the parabola at the point where x = -5, we need to first find the derivative of the function y = 4x^2 - 3x + 5.
The derivative of a quadratic function can be found by applying the power rule, which states that for a function of the form f(x) = ax^n, where a is a constant and n is a positive integer, the derivative is f'(x) = nax^(n-1).
In this case, our function is y = 4x^2 - 3x + 5, and we want to find the derivative with respect to x. Applying the power rule, we get:
dy/dx = d/dx(4x^2) - d/dx(3x) + d/dx(5)
= 8x - 3
Now that we have the derivative, we can evaluate it at x = -5 to find the slope of the tangent line at that point. Plugging in x = -5, we get:
m = 8(-5) - 3
= -40 - 3
= -43
So, the slope of the tangent line is -43.
Now, to write the equation of the tangent line in the form y = mx + b, we need to find the y-intercept b. We already know the slope m, which is -43, and we can use the point-slope form of a linear equation to find b.
The point-slope form is given by:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line. In this case, we have the point (-5, y), where y is the y-coordinate of the point on the parabola corresponding to x = -5.
Plugging in the values, we have:
y - y1 = -43(x - x1)
y - y = -43(x - (-5))
y = -43(x + 5)
Therefore, the equation of the tangent line can be written as y = -43x - 215, where b is -215.