Find an equation of the tangent to the curve f(x) = 2x^2 - 2x + 1 that has slope 2.
f' = slope = 4 x -2
so
4x-2 = 2
4 x = 4
x = 1
if x = 1 then
y = 2 -2 + 1 = 1
so tangent at point (1,1) and m = 2
1 = 2 (1) + b
b = -1
so
y = 2 x - 1
Well, I must say, finding the equation of the tangent can be a bit "tangential" itself! But fear not, I am here to help!
To find the equation of the tangent line with a slope of 2, we need to use a little math magic.
First, let's find the derivative of the function f(x) = 2x^2 - 2x + 1. Taking the derivative will give us the slope of the tangent line at any given point on the curve.
The derivative of f(x) = 2x^2 - 2x + 1 is f'(x) = 4x - 2.
Now, we want the slope of the tangent line to be 2, so let's set f'(x) equal to 2 and solve for x:
4x - 2 = 2.
Adding 2 to both sides, we get:
4x = 4.
Dividing both sides by 4, we find:
x = 1.
So, the x-coordinate of the point where the tangent line touches the curve is x = 1. Plugging this value back into the original function, we find:
f(1) = 2(1)^2 - 2(1) + 1 = 2 - 2 + 1 = 1.
Therefore, the point of tangency is (1, 1).
Now that we have the point of tangency, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is y - y1 = m(x - x1), where (x1, y1) is the point and m is the slope.
Substituting the values we found, the equation of the tangent line is:
y - 1 = 2(x - 1).
Simplifying, we get:
y - 1 = 2x - 2.
Or, if you prefer:
2x - y = 1.
And there you have it! The equation of the tangent line to the curve f(x) = 2x^2 - 2x + 1 with a slope of 2 is 2x - y = 1. Keep those tangents rolling!
To find an equation of the tangent line to the curve f(x) = 2x^2 - 2x + 1 that has slope 2, we need to find the x-coordinate of the point where the tangent line touches the curve.
Step 1: Find the derivative of f(x) to get the expression for the slope of the tangent line.
The derivative of f(x) is given by f'(x) = 4x - 2.
Step 2: Since we know the slope of the tangent line is 2, we set f'(x) = 2 and solve for x.
Setting 4x - 2 = 2, we have:
4x = 4
x = 1.
Step 3: Now that we have the x-coordinate of the point of tangency, we can find the y-coordinate by substituting x = 1 into the equation of the curve.
f(1) = 2(1)^2 - 2(1) + 1
f(1) = 2 - 2 + 1
f(1) = 1.
Therefore, the point of tangency is (1, 1).
Step 4: Now we have the slope of the tangent line (2) and the point of tangency (1, 1). We can use the point-slope form of a linear equation to find the equation of the tangent line.
The point-slope form of a linear equation is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.
Substituting the values into the point-slope form, we get:
y - 1 = 2(x - 1)
y - 1 = 2x - 2
y = 2x - 1.
Therefore, the equation of the tangent line to the curve f(x) = 2x^2 - 2x + 1 that has a slope of 2 is y = 2x - 1.
To find the equation of a tangent to the curve, we need two pieces of information: a point on the curve and the slope of the tangent line.
First, let's find the points on the curve that satisfy the condition of having a slope of 2. The slope of a function at any given point is given by the derivative of the function.
To find the derivative of f(x) = 2x^2 - 2x + 1, we can apply the power rule of differentiation. Taking the derivative of each term individually:
f'(x) = d/dx (2x^2) - d/dx (2x) + d/dx (1)
Using the power rule, the derivatives of the terms are:
f'(x) = 4x - 2
Now, we set the derivative equal to 2 to find the x-values where the slope is 2:
4x - 2 = 2
Adding 2 to both sides:
4x = 4
Dividing both sides by 4:
x = 1
So, the point on the curve where the slope is 2 is (1, f(1)).
Now that we have the point (1, f(1)) and the slope (2), we can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1)
where (x1, y1) is the point on the curve and m is the slope of the tangent line. Plugging in the values:
y - f(1) = 2(x - 1)
Expanding and simplifying:
y - f(1) = 2x - 2
To express the equation in the slope-intercept form (y = mx + b), we can add f(1) to both sides:
y = 2x + (f(1) - 2)
So, the equation of the tangent to the curve f(x) = 2x^2 - 2x + 1 that has slope 2 is y = 2x + (f(1) - 2), where f(1) is the value of the function at x = 1.