If G(x) = x^2 – 3x + 3, find G'(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). y1(x) y2(x)

First, we need to find the derivative of G(x), which will give us the slope of the tangent line at any point on the curve: G'(x) = 2x - 3 Now, we can use this to find the equations of the tangent lines to the curve at the points (0, 3) and (4, 7): At (0, 3): G'(0) = 2(0) - 3 = -3 So the slope of the tangent line at (0, 3) is -3. Using point-slope form, we can write the equation of the tangent line: y - 3 = -3(x - 0) Simplifying, we get: y = -3x + 3 So the equation of the tangent line to the curve at (0, 3) is y1(x) = -3x + 3. At (4, 7): G'(4) = 2(4) - 3 = 5 So the slope of the tangent line at (4, 7) is 5. Using point-slope form, we can write the equation of the tangent line: y - 7 = 5(x - 4) Simplifying, we get: y = 5x - 13 So the equation of the tangent line to the curve at (4, 7) is y2(x) = 5x - 13.