Find the equation of the tangent line to the curve y = 2x^2 - 4x + 3 at the point (3,9)

To find the equation of the tangent line to the curve y = 2x² - 4x + 3, we have to differentiate. Taking each part of the equation separately, we can calculate dy/dx = 4x - 4. By differentiating the equation for the curve, we can calculate the gradient of the curve at any point. We have been given the point (3,9), and so we can substitute x = 3 into dy/dx = 4x - 4 to find the gradient of the curve at that point; dy/dx = 12 - 4, therefore dy/dx = 8.We have found the gradient of the curve to be 8 at the point (3,9) - as we are trying to find the equation of the tangent line to the curve, which has the same gradient as the curve, we know that the gradient of the tangent is also 8.From here we can find the equation of the tangent, which will be in the format y = mx + c where m is the gradient of the line and c is the y-intercept. As we have calculated the gradient of the tangent to be 8, y = 8x + c. We now have to substitute the point (3,9) into this equation (as we know that the tangent line passes through this point); 9 = 8(3) + c, 9 = 24 + c; and therefore c = 9 -24, c = -15. The equation of the tangent line to the curve y = 2x² - 4x + 3 is therefore y = 8x - 15