find the gradient of the tangent to the curve y=3x^2+2x-1 p(1,4)
Slope = gradient if tangent line = dy/dx = m
In this case:
dy / dx = m = 3 • 2 x + 2 = 6 x + 2
In point where x = 1
m = 6 x + 2 = 6 • 1 + 2 = 6 + 2 = 8
Point slope equation of straight line is:
y − y1 = m ( x − x1 )
In this case:
x1 = 1 , y1 = 4 , m = 8
so
y − y1 = m ( x − x1 )
y − 4 = 8 ( x − 1 )
y − 4 = 8 x − 8
Add 4 to both sides.
y = 8 x - 4
My typo.
Slope = gradient of tangent line = dy/dx = m
To find the gradient of the tangent to the curve at a certain point, you need to find the derivative of the curve and evaluate it at the given point.
The given curve is y = 3x^2 + 2x - 1.
To find the derivative, you'll need to apply the power rule and sum/difference rule of differentiation. The power rule states that for any power function f(x) = x^n, the derivative is f'(x) = nx^(n-1). And for a sum or difference of functions, the derivative of the sum (or difference) is simply the sum (or difference) of the derivatives.
Differentiating y = 3x^2 + 2x - 1:
dy/dx = d(3x^2)/dx + d(2x)/dx + d(-1)/dx
Using the power rule, we get:
dy/dx = 6x + 2 + 0
Simplifying, we have:
dy/dx = 6x + 2
Now, we plug in the x-coordinate of the given point, which is x = 1, to find the gradient (or slope) at that point:
Gradient = dy/dx at x = 1
= 6(1) + 2
= 6 + 2
= 8
Therefore, the gradient of the tangent to the curve y = 3x^2 + 2x - 1 at the point (1,4) is 8.
To find the gradient of the tangent to the curve, you need to find the derivative of the given function. The derivative represents the slope of the tangent line to the curve at any given point.
Step 1: Start with the given equation of the curve, y = 3x^2 + 2x - 1.
Step 2: Differentiate the equation with respect to x using the power rule. Each term in the equation is a polynomial, and for each term, you need to multiply the coefficient by the power of x and reduce the power by 1.
Differentiating the equation, we get:
dy/dx = d/dx (3x^2) + d/dx (2x) - d/dx (1)
dy/dx = 6x + 2 - 0
dy/dx = 6x + 2
Step 3: Now you have the derivative equation, dy/dx = 6x + 2. This represents the slope of the tangent line at any point on the curve.
Step 4: To find the gradient (slope) of the tangent at a specific point, plug in the x-coordinate of the point into the derivative equation. In this case, the given point is P(1,4).
Substitute x = 1 into the derivative equation:
dy/dx = 6(1) + 2
dy/dx = 6 + 2
dy/dx = 8
Therefore, at the point P(1,4), the gradient (slope) of the tangent to the curve y = 3x^2 + 2x - 1 is 8.