The slope of the tangent line to the parabola y=4x2–3x+5 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:______?

The first to blanks are -43, but what is b? thanks

User Icon for bobpursley bobpursley answered
14 years ago

y'(-5)= 8(-5)-3=-43

now for the line
y= mx+b
where x,y is a point on the curve
x=-5, so y= 4x^2-3x+5 solve for the y.
Now that you have x,y, put it in the equation y=mx+b and solve for b.

User Icon for Step-by-Step Bot Step-by-Step Bot answered
1 year ago

To find the slope of the tangent line to the parabola y = 4x^2 - 3x + 5 at the point where x = -5, we need to find the derivative of the function and evaluate it at x = -5.

Step 1: Find the derivative of the function y = 4x^2 - 3x + 5.
To find the derivative, we differentiate each term separately using the power rule.
The derivative of 4x^2 is 8x.
The derivative of -3x is -3.
The derivative of 5 is 0.

So, dy/dx = 8x - 3.

Step 2: Evaluate the derivative at x = -5.
Substitute x = -5 into the derivative equation:
dy/dx = 8(-5) - 3
dy/dx = -40 - 3
dy/dx = -43.

Therefore, the slope of the tangent line to the parabola at x = -5 is -43.

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.

Step 1: Use the point-slope form of a line.
The point-slope form of a line is y - y1 = m(x - x1).
Using the given point (x1, y1) = (-5, 4(-5)^2 - 3(-5) + 5) = (-5, 4(25) + 15 + 5) = (-5, 100 + 15 + 5) = (-5, 120), we can write the equation as:
y - 120 = -43(x - (-5)).

Step 2: Simplify the equation.
y - 120 = -43(x + 5)
y - 120 = -43x - 215
y = -43x - 215 + 120
y = -43x - 95.

Therefore, the equation of the tangent line can be written as y = -43x - 95, where m is -43 and b is -95.

User Icon for Explain Bot Explain Bot answered
11 months ago

To find the slope of the tangent line, we need to determine the derivative of the parabola at the given point. The equation of the parabola is y = 4x^2 - 3x + 5.

First, let's find the derivative of the parabola using the power rule of differentiation. Differentiating the terms of the equation one by one:

dy/dx = d/dx(4x^2 - 3x + 5)
= d/dx(4x^2) - d/dx(3x) + d/dx(5)
= 8x - 3

Now that we have the derivative, we can find the slope of the tangent line at x = -5 by substituting -5 into the derivative equation:

m = dy/dx = 8(-5) - 3
= -40 - 3
= -43

Therefore, the slope of the tangent line is -43.

To find the equation of the tangent line in the form y = mx + b, we have the slope (m) as -43. Now, we need to find the y-intercept (b).

To find b, we substitute the point (x, y) = (-5, f(-5)) into the original equation of the parabola, y = 4x^2 - 3x + 5, and solve for y:

y = 4(-5)^2 - 3(-5) + 5
= 4(25) + 15 + 5
= 100 + 15 + 5
= 120

Therefore, the point of tangency is (-5, 120). Now we can substitute this point, along with the slope (m = -43), into the equation y = mx + b and solve for b:

120 = -43(-5) + b
120 = 215 + b
120 - 215 = b
-95 = b

Therefore, b is -95.

Hence, the equation of the tangent line is y = -43x - 95.