(a) if ( g(x)=x^{2}-3x + 3 ), find ( g^{prime}(a) ) and use it to find equations of the tangent lines to the…

(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.

(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.

Answer

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)