We have to find the tangent line to y = x² - 3 at x = a.
The slope of the tangent line will be equal to the first derivative at that point.
We start by calculating the first derivative using the definition of the derivative:
[tex]\begin{gathered} y^{\prime}=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \\ y^{\prime}=\lim_{h\to0}\frac{(x+h)^2-3-x^2+3}{h} \\ y^{\prime}=\lim_{h\to0}\frac{x^2+2xh-h^2-x^2-3+3}{h} \\ y^{\prime}=\lim_{h\to0}\frac{2xh-h^2}{h} \\ y^{\prime}=\lim_{h\to0}2x-h \\ y^{\prime}=2x \end{gathered}[/tex]
As we have to calculate the tangent line at x = 2 and a = 2, we can replace x with 2 to find the slope of the tangent line:
[tex]m=y^{\prime}(2)=2(2)=4[/tex]
We now need a point of the line to find its complete equation.
As the line is tangent to (a, f(a)) we can calculate the value of f(a) = f(2) as:
[tex]f(2)=(2)^2-3=4-3=1[/tex]
Then, with a slope m = 4 and tangent to point (2, 1) we can write the equation of the line as:
[tex]\begin{gathered} y-y_0=m(x-x_0) \\ y-1=4(x-2) \\ y=4x-8+1 \\ y=4x-7 \end{gathered}[/tex]
We can check this with a graph as:
Answer:
The tangent line is y = 4x - 7
