Answer:
The equation of a tangent line is y = 4x-7
Step-by-step explanation:
We are given our quadratic function:-
[tex] \displaystyle \large{f(x) = {x}^{2} - 3}[/tex]
Since we want to find a line that is tangent to the parabola, let's recall our basic differential.
Differential (Power/Exponent)
[tex] \displaystyle \large{f(x) = a {x}^{n} \longrightarrow f'(x) = na {x}^{n - 1} }[/tex]
Differential (Constant)
[tex] \displaystyle \large{f(x) = a \longrightarrow f'(x) = 0}[/tex]
For a = constant.
First, differentiate the function.
[tex] \displaystyle \large{f(x) = {x}^{2} - 3} \\ \displaystyle \large{f'(x) = 2{x}^{2 - 1} - 0} \\ \displaystyle \large{f'(x) = 2x}[/tex]
Then substitute x = 2 in f'(x) to find the slope at x = 2 for parabola.
[tex] \displaystyle \large{f'(2)= 2(2)} \\ \displaystyle \large{f'(2)= 4}[/tex]
Therefore, slope at x = 2 is 4.
Form a point-slope equation:-
Point-Slope (Derivative)
[tex] \displaystyle \large{f(x) - f(a) = f'(a)(x - a)}[/tex]
Let a = x = 2
To find f(a), substitute x = a = 2 in x^2-3.
[tex] \displaystyle \large{f(2) = {2}^{2} - 3} \\ \displaystyle \large{f(2) = 4- 3} \\ \displaystyle \large{f(2) = 1}[/tex]
Therefore our f(a) is 1.
We know:-
Therefore the tangent equation is:-
[tex] \displaystyle \large{f(x) - 1 = 4(x - 2)} \\ \displaystyle \large{f(x) = 4(x - 2) + 1} \\ \displaystyle \large{f(x) = 4x - 8 + 1} \\ \displaystyle \large{f(x) = 4x - 7}[/tex]
