The topic is the application of function derivatives. The first task: Find the equation of the tangent line to 𝑦=2𝑥2−𝑥+3 at 𝑥0=0.5 Give explanations and formulas and draw a graph if needed

Sure, here is the step-by-step solution:

1. Find the derivative of the function: The derivative of a function gives us the slope of the tangent line at any point. The derivative of 𝑦=2𝑥²−𝑥+3 is obtained using the power rule (nx^(n-1)) for differentiation. So, 𝑦'=4𝑥-1.

2. Evaluate the derivative at 𝑥₀=0.5: Substituting 𝑥₀ into 𝑦' gives us the slope of the tangent line at 𝑥₀. So, 𝑦'(0.5)=4(0.5)-1=1.

3. Find the y-coordinate of the tangent point: Substituting 𝑥₀ into the original function 𝑦 gives us the y-coordinate of the point where the tangent line touches the curve. So, 𝑦(0.5)=2(0.5)²-(0.5)+3=2.5.

4. Use the point-slope form to find the equation of the tangent line: The point-slope form of a line is 𝑦−𝑦₁=𝑚(𝑥−𝑥₁), where (𝑥₁, 𝑦₁) is a point on the line and 𝑚 is the slope of the line. Substituting 𝑚=1, 𝑥₁=0.5, and 𝑦₁=2.5 gives us the equation of the tangent line: 𝑦−2.5=1(𝑥−0.5), or 𝑦=𝑥+2.

As for the graph, it would show the function 𝑦=2𝑥²−𝑥+3, which is a parabola opening upwards, and the line 𝑦=𝑥+2, which is a straight line. The line touches the parabola at the point (0.5, 2.5).