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
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.
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.
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.