Questions: Differentiation Formative assessment [40 marks] Your name. MagyR 3. Consider the curve y=x^2-4 x+2. (a) Find an expression for dy/dx. (b) Show that the normal to the curve at the point where x=1 is 2 y-x+3=0.

Solution Steps

(a) To find an expression for \(\frac{dy}{dx}\), differentiate the given curve equation \(y = x^2 - 4x + 2\) with respect to \(x\).

(b) To show that the normal to the curve at the point where \(x=1\) is \(2y - x + 3 = 0\), first find the derivative \(\frac{dy}{dx}\) to get the slope of the tangent at \(x=1\). Then, use the negative reciprocal of this slope to find the slope of the normal. Finally, use the point-slope form of a line to find the equation of the normal and verify it matches the given equation.

Given the curve \( y = x^2 - 4x + 2 \), we differentiate it with respect to \( x \):

\[ \frac{dy}{dx} = 2x - 4 \]

Substituting \( x = 1 \) into the derivative to find the slope of the tangent:

\[ \text{slope}_{\text{tangent}} = 2(1) - 4 = -2 \]

The slope of the normal line is the negative reciprocal of the slope of the tangent:

\[ \text{slope}_{\text{normal}} = -\frac{1}{\text{slope}_{\text{tangent}}} = -\frac{1}{-2} = \frac{1}{2} \]

Now, we calculate the \( y \)-coordinate at \( x = 1 \):

\[ y(1) = 1^2 - 4(1) + 2 = -1 \]

Thus, the point on the curve is \( (1, -1) \).

Using the point-slope form of the line, the equation of the normal line is:

\[ y - (-1) = \frac{1}{2}(x - 1) \]

Simplifying this gives:

\[ y + 1 = \frac{1}{2}x - \frac{1}{2} \]

Rearranging leads to:

\[ 2y - x + 3 = 0 \]

Final Answer