Find the equation of the tangent line at point (2, 2) on the curve 𝑓(𝑥)=𝑥2−7𝑥+12. Bold text start[5 marks]

Sure, here are the steps to find the equation of the tangent line at point (2, 2) on the curve 𝑓(𝑥)=𝑥²−7𝑥+12.

Step 1: Find the derivative of the function 𝑓(𝑥)=𝑥²−7𝑥+12. The derivative of a function gives us the slope of the tangent line at any point on the curve.

The derivative of 𝑥² is 2𝑥, the derivative of -7𝑥 is -7, and the derivative of a constant like 12 is 0. So, the derivative of 𝑓(𝑥) is 𝑓'(𝑥) = 2𝑥 - 7.

Step 2: Substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line at that point.

Substituting 2 for 𝑥 in 𝑓'(𝑥) = 2𝑥 - 7 gives us 𝑓'(2) = 2*2 - 7 = -3. So, the slope of the tangent line at the point (2, 2) is -3.

Step 3: Use the point-slope form of a line, y - y₁ = m(x - x₁), to find the equation of the tangent line. Here, m is the slope of the line, and (x₁, y₁) is a point on the line.

Substituting -3 for m, 2 for x₁, and 2 for y₁ in the equation y - y₁ = m(x - x₁) gives us y - 2 = -3(x - 2).

Step 4: Simplify the equation to put it in slope-intercept form, y = mx + b.

Distributing -3 in the equation y - 2 = -3(x - 2) gives us y - 2 = -3x + 6. Adding 2 to both sides of the equation to solve for y gives us y = -3x + 8.

So, the equation of the tangent line at point (2, 2) on the curve 𝑓(𝑥)=𝑥²−7𝑥+12 is y = -3x + 8.