Question

(a) if ( g(x)=x^{2}-3x + 3 ), find ( g^{prime}(a) ) and use it to find equations of the tangent lines to the curve ( y=x^{2}-3x + 3 ) at the points ( (0,3) ) and ( (4,7) ). ( g^{prime}(a)= ) ( y_{1}(x)= ) (passing through ( (0,3) )) ( y_{2}(x)= ) (passing through ( (4,7) )) (b) illustrate part (a) by graphing the curve and the tangent lines on the same screen.

Explanation:

Step1: Find the derivative of \(G(x)\)

Using the power rule \((x^n)^\prime = nx^{n - 1}\), for \(G(x)=x^{2}-3x + 3\), we have \(G^\prime(x)=\frac{d}{dx}(x^{2})-\frac{d}{dx}(3x)+\frac{d}{dx}(3)\).
Since \(\frac{d}{dx}(x^{2}) = 2x\), \(\frac{d}{dx}(3x)=3\), and \(\frac{d}{dx}(3) = 0\), then \(G^\prime(x)=2x - 3\). So \(G^\prime(a)=2a-3\).

Step2: Find the slope of the tangent line at \((0,3)\)

Substitute \(x = 0\) into \(G^\prime(x)\). The slope \(m_1=G^\prime(0)=2\times0 - 3=-3\).
Using the point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(0,3)\) and \(m=-3\), we get \(y-3=-3(x - 0)\), which simplifies to \(y_1(x)=-3x + 3\).

Step3: Find the slope of the tangent line at \((4,7)\)

Substitute \(x = 4\) into \(G^\prime(x)\). The slope \(m_2=G^\prime(4)=2\times4-3=5\).
Using the point - slope form \(y - y_1=m(x - x_1)\) with \((x_1,y_1)=(4,7)\) and \(m = 5\), we have \(y - 7=5(x - 4)\).
Expand: \(y-7=5x-20\), then \(y_2(x)=5x-13\).

Answer:

\(G^\prime(a)=2a - 3\)
\(y_1(x)=-3x + 3\)
\(y_2(x)=5x-13\)