Find the slope of the normal to the curve y=4x²-14x+5 at x=5

The correct option is – (1/26).

To find the slope of the normal to the curve, we first need to determine the slope of the tangent to the curve at the given point, and then use the relationship between the slope of a tangent and the slope of a normal.

Step 1: Find the derivative of the curve's equation
The given equation of the curve is:
$$y=4x^2-14x+5$$
Differentiating both sides with respect to $x$ gives the rate of change, which represents the slope of the tangent ($\frac{dy}{dx}$):
$$\frac{dy}{dx}=\frac{d}{dx}(4x^2-14x+5)$$
Applying the power rule of differentiation:
$$\frac{dy}{dx}=8x-14$$

Step 2: Calculate the slope of the tangent at $x$ = 5
Substitute $x=5$ into the derivative:
$${\left(\frac{dy}{dx}\right)}_{x=5}=8\left(5\right)-14$$
$${\left(\frac{dy}{dx}\right)}_{x=5}=40-14=26$$
So, the slope of the tangent ($m_t$) at $x=5$ is 26.

Step 3: Determine the slope of the normal
Since the normal line is perpendicular to the tangent line, the relationship between their slopes is:
$$m_n=-\frac{1}{m_t}$$
Substituting the value of the slope of the tangent:
$$m_n=-\frac{1}{26}$$
Thus, the slope of the normal to the curve at $x=5$ is $-\frac{1}{26}$, which corresponds to the option – (1/26).