How many critical points can a quadratic polynomial function have?
Answer
604.8k+ views
Hint: In this question, they have asked us to find the number of critical points that a quadratic polynomial equation can have. Critical point is said to be the point where $\dfrac{{dy}}{{dx}} = 0$ . Therefore we need to assume an equation and then find its derivative to check how many critical points exist for a quadratic polynomial equation.
Complete step-by-step solution:
Here, we need to find how many critical points a quadratic polynomial equation can have.
Critical point is said to be a point or the roots of the given equation where the first derivative is equal to zero that is $\dfrac{{dy}}{{dx}} = 0$
Let us assume a quadratic polynomial equation first,
$y = a{x^2} + bx + c$, where $a \ne 0$
We need to find the critical point; hence we have to find the first derivative of the equation.
Derivation of $a{x^2} + bx + c$
$ \Rightarrow \dfrac{d}{{dx}}a{x^2} + bx + c = 2ax + b$
Now we get,
\[ \Rightarrow 2ax + b\]
As this is first degree, we need to find the value of $x$ in $\dfrac{{dy}}{{dx}} = 0$
\[ \Rightarrow 2ax + b = 0\]
Transferring the variable $b$ to the other side, we get
$ \Rightarrow 2ax = - b$
Finding the value of $x$,
$ \Rightarrow x = \dfrac{{ - b}}{{2a}}$
Here, we got a single solution for $x$
Since it has a single solution, a quadratic polynomial function can have a single critical point.
Note: A critical point of a continuous function is a point at which the derivative is zero or undefined. Critical points are the points on the graph where the function's rate of change is altered by either a change from increasing to decreasing. If our equation is \[f\left( x \right) = mx + b\] , we get\[f'\left( x \right) = m\] . So if the function is constant \[\left( {m = 0} \right)\] we get infinitely many critical points. Otherwise, we have no critical points.
Complete step-by-step solution:
Here, we need to find how many critical points a quadratic polynomial equation can have.
Critical point is said to be a point or the roots of the given equation where the first derivative is equal to zero that is $\dfrac{{dy}}{{dx}} = 0$
Let us assume a quadratic polynomial equation first,
$y = a{x^2} + bx + c$, where $a \ne 0$
We need to find the critical point; hence we have to find the first derivative of the equation.
Derivation of $a{x^2} + bx + c$
$ \Rightarrow \dfrac{d}{{dx}}a{x^2} + bx + c = 2ax + b$
Now we get,
\[ \Rightarrow 2ax + b\]
As this is first degree, we need to find the value of $x$ in $\dfrac{{dy}}{{dx}} = 0$
\[ \Rightarrow 2ax + b = 0\]
Transferring the variable $b$ to the other side, we get
$ \Rightarrow 2ax = - b$
Finding the value of $x$,
$ \Rightarrow x = \dfrac{{ - b}}{{2a}}$
Here, we got a single solution for $x$
Since it has a single solution, a quadratic polynomial function can have a single critical point.
Note: A critical point of a continuous function is a point at which the derivative is zero or undefined. Critical points are the points on the graph where the function's rate of change is altered by either a change from increasing to decreasing. If our equation is \[f\left( x \right) = mx + b\] , we get\[f'\left( x \right) = m\] . So if the function is constant \[\left( {m = 0} \right)\] we get infinitely many critical points. Otherwise, we have no critical points.
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