# Find (a) the slope of the curve at the given point (P) and (b) an equation of the tangent line at P.

y = 5 - 6x^{2}; P(2, 19)

**Solution:**

a) To find the slope of the curve, the derivative of a function f(x) at the given point is a slope of the curve

Slope of the given curve y = 5 - 6x^{2} at the point P(2, 19) is the derivative f'(x) at P(2, 19)

m = f'(x) = dy/dx

= 0 - 6(2x)

Since the derivative of a constant is zero and the derivative of x^{2} is 2x

f'(x) at P(2, 19) is 0 - 6(2 × 2)

= 0 - 6 × 4 = -24

b) To find an equation of the tangent line at P (2, 19) for the curve y = 5 - 6x^{2}

The equation of tangent line in for the function f(x) at P (x_{1}, y_{1}) is (y - y_{1}) = m (x - x_{1})

Given:

x_{1} = 2 and y_{1} = 19, f(x) = 5 - 6x^{2} and slope of tangent m = -24

The equation of the tangent line is (y - 19) = - 24(x - 2)

⇒ y - 19 = -24x + 48

⇒ 24x + y = 48 + 19

The equation of tangent is 24x + y = 67

## Find (a) the slope of the curve at the given point (P), and (b) an equation of the tangent line at P. y= 5 - 6x²; P (2, 19)

**Summary:**

The equation of tangent is 24x + y = 67 with the slope m = -24. Application of slope of tangent is temperature change at a particular time, velocity of a falling object at a particular time, current through a circuit at a particular time, variation in stock market prices at a particular time, population growth at a particular time, temperature increases as density increases in a gas.

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