Question

Use the definition of the derivative to prove that every quadratic function $f\left(x\right)=a{x}^2+bx+c$ has the property that its graph has a horizontal tangent line at the point $x=-\frac{b}{2a}$ .

Short answer

By using the concept that horizontal tangent line is a mathematical figure on a graph , located where a function's derivative is zero, we find that a quadratic function$f\left(x\right)=a{x}^2+bx+c$has a horizontal tangent line at a point$x=-\frac{b}{2a}$

Step-by-step solution

Step 1. Given Information

We have given the following quadratic function $f\left(x\right)=a{x}^2+bx+c$.

We have to prove that it has a horizontal tangent line at point $x=-\frac{b}{2a}$.

We know that a horizontal tangent line of a function is located where first derivative is zero.

We find the point where the first derivative of given function is zero to complete our proof.

Step 2. To find derivative of function

The given quadratic function is :-

$f\left(x\right)=a{x}^2+bx+c$

Now we know that the derivative of function $f\left(x\right)$is :-

role="math" $\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$

Put the values :-

role="math" $$\begin{gathered} \underset{h\to 0}{\lim }\frac{[a{(x+h)}^2+b(x+h)+c]-[a{x}^2+bx+c]}{h} \\ =\underset{h\to 0}{\lim }\frac{\left[a\right(x^2+h^2+2xh)+bx+bh+c]-[a{x}^2+bx+c]}{h} \\ =\underset{h\to 0}{\lim }\frac{a{x}^2+a{h}^2+2axh+bx+bh+c-a{x}^2-bx-c}{h} \\ =\underset{h\to 0}{\lim }\frac{a{h}^2+2axh+bh}{h} \\ =\underset{h\to 0}{\lim }\frac{h(ah+2ax+b)}{h} \\ =\underset{h\to 0}{\lim }ah+2ax+b \\ =a\left(0\right)+2ax+b \\ =2ax+b \end{gathered}$$

So the derivative of a given quadratic function isrole="math" localid="1648375882379" $2ax+b$.

Step 3. Take derivative equals to zero

We find that the derivative of given quadratic function $a{x}^2+bx+c$is $2ax+b$.

Now take the derivative equals to zero, then we have :-

$2ax+b=0$.

Solve the equation for $x$:-

role="math" $$\begin{gathered} 2ax+b=0 \\ \Rightarrow 2ax=-b \\ \Rightarrow x=-\frac{b}{2a} \end{gathered}$$

This point is same as the given point.

So it is proved that horizontal tangent of any quadratic function$a{x}^2+bx+c$is at point$x=-\frac{b}{2a}$.