Question

Identify the coordinates of any local and absolute extreme points and inflection points of the following functions. Graph the function.
(A) $y={{x}^{2}}-4x+3$
(B) $y={{x}^{2}}-3x+3$

Answer 1

Add and subtract 1 in the expression $y={{x}^{2}}-4x+3$ and modify the expression into a perfect square form. The maximum value of the square term is infinity and the minimum value of the square term is equal to zero. Now, make the square term equal to zero and calculate the value of x. Similarly, add and subtract $\dfrac{9}{4}$ in the expression $y={{x}^{2}}-3x+3$ and modify the expression into a perfect square form. The maximum value of the square term is infinity and the minimum value of the square term is equal to zero. Now, make the square term equal to zero and calculate the value of x. We know the property that the points of inflection are such points where the concavity of the function changes. Now, check whether the points of inflection exist or not for the given functions.

Complete answer:
According to the question, we are given two functions which are
$y={{x}^{2}}-4x+3$ ……………………………………(1)
$y={{x}^{2}}-3x+3$ …………………………………….(2)
First of all, let us try to solve equation (1).
On adding and subtracting 1 in equation (1), we get

& \Rightarrow y={{x}^{2}}-4x+3+1-1 \\\ & \Rightarrow y={{x}^{2}}-4x+4-1 \\\ \end{aligned}$$ $$\Rightarrow y={{\left( x-2 \right)}^{2}}-1$$ …………………………………….(3) Since the expression in the equation has one square term and we know that the square term is always greater than or equal to zero so, the maximum value of the square term will tend to infinity. Therefore, when x will tend to infinity the value of the expression shown in equation (3) will also tend to infinity ……………………………………….(4) But, for its minimum value, the square term must be equal to zero i.e., $$\Rightarrow {{\left( x-2 \right)}^{2}}=0$$ $$\Rightarrow x=2$$ The maximum value of the expression $$y={{x}^{2}}-4x+3$$ tends to infinity when x tends to infinity …………………………….(5) The expression, $$y={{x}^{2}}-4x+3$$ poses its minimum value when $$x=2$$ …………………………….(6) Now, let us solve equation (2). On adding and subtracting $$\dfrac{9}{4}$$ in equation (2), we get $$\begin{aligned} & \Rightarrow y={{x}^{2}}-3x+3+\dfrac{9}{4}-\dfrac{9}{4} \\\ & \Rightarrow y={{x}^{2}}-2\times x\times \dfrac{3}{2}+\dfrac{9}{4}+3-\dfrac{9}{4} \\\ & \Rightarrow y={{\left( x-\dfrac{3}{2} \right)}^{2}}+\dfrac{12-9}{4} \\\ \end{aligned}$$ $$\Rightarrow y={{\left( x-\dfrac{3}{2} \right)}^{2}}+\dfrac{3}{4}$$ …………………………………………(7) Since the expression in the equation has one square term and we know that the square term is always greater than or equal to zero so, the maximum value of the square term will tend to infinity. Therefore, when x will tend to infinity the value of the expression shown in equation (7) will also tend to infinity ……………………………………….(8) But, for its minimum value, the square term must be equal to zero i.e., $$\Rightarrow {{\left( x-\dfrac{3}{2} \right)}^{2}}=0$$ $$\Rightarrow x=\dfrac{3}{2}$$ The maximum value of the expression, $$y={{x}^{2}}-3x+3$$ tends to infinity when x tends to infinity …………………………….(9) The expression, $$y={{x}^{2}}-3x+3$$ poses its minimum value when $$x=\dfrac{3}{2}$$ …………………………….(10) From equation (5) and equation (6), we have the coordinates for the extremum values of the expression, $${{x}^{2}}-4x+3$$ . So, the extremum value of the expression $$y={{x}^{2}}-4x+3$$ will be at $$x=2$$ and $$x\to \infty $$ . Similarly, from equation (9) and equation (10), we have the coordinates for the extremum values of the expression, $${{x}^{2}}-3x+3$$ . So, the extremum value of the expression $$y={{x}^{2}}-3x+3$$ will be at $$x=\dfrac{3}{2}$$ and $$x\to \infty $$ . We also know the property that the points of inflection are such points where the concavity of the function changes. Since a quadratic equation can be either concave upward or concave down. It doesn’t have any change of concavity. Therefore, there will no any points of inflection for the quadratic equation …………………………………………(11) Using the property shown in equation (11), we can say that the expressions $$y={{x}^{2}}-4x+3$$ and $$y={{x}^{2}}-3x+3$$ don’t have the points of inflection. Now, on plotting the graph of $$y={{x}^{2}}-4x+3$$ , we get  Similarly, on plotting the graph of $$y={{x}^{2}}-3x+3$$ , we get  Therefore, the coordinates of the local and absolute extreme points and inflection points of the given functions have been calculated. **Note:** Whenever we have the question where a quadratic equation is given and we have to find its extremum values or coordinates for the extremum values. Always approach this type of question by modifying the given quadratic equation as per the perfect square expression. Modifying this form will make it easy to comment on extremum points for the quadratic equation.