Question

How do you graph $y={{x}^{2}}-2x+3$?

Answer 1

To solve equation we will first check coefficient of ${{x}^{2}}$ and use formula $c-\dfrac{{{b}^{2}}}{4a}$ for finding its minimum value. Also, we will use formulas, ${{x}^{2}}=4ay$, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and $x=\dfrac{-b}{2a}$ along with substitution as x = 0, x = -1, x = 1 and so on. After this we will collect these points and plot these on the graph. At last we will join these points together and get the required graph.

Complete step by step solution:
To draw the graph of a quadratic function we will use some methods so that we can find some points and plot them on the graph.
We will find minimum and maximum points of a quadratic function. For this we need to check for the coefficient of ${{x}^{2}}$. In the equation $y={{x}^{2}}-2x+3$, we have the coefficient of ${{x}^{2}}$ as 1. Since, it is greater than 0 therefore, we come to the conclusion of its maximum point as infinity or a number which is maximum here but is leading to infinity.
Also, we come to know that the given equation will only have a minimum point so, for finding it we will use the formula $c-\dfrac{{{b}^{2}}}{4a}$. By equating general equation $a{{x}^{2}}+2x+c=0$ to the given equation $y={{x}^{2}}-2x+3$ we get a = 1, b = - 2 and c = 3. Therefore, by substituting it in the formula $c-\dfrac{{{b}^{2}}}{4a}$ we get,

& c-\dfrac{{{b}^{2}}}{4a}=3-\dfrac{{{\left( -2 \right)}^{2}}}{4\left( 1 \right)} \\\ & \Rightarrow c-\dfrac{{{b}^{2}}}{4a}=3-\dfrac{4}{4} \\\ & \Rightarrow c-\dfrac{{{b}^{2}}}{4a}=3-1=2...(i) \\\ \end{aligned}$$ Thus, 2 will be the minimum value of equation $y={{x}^{2}}-2x+3$. Now, we will compare general equation of parabola ${{x}^{2}}=4ay\,\,\text{Or,}\,\,y=\dfrac{{{x}^{2}}}{4a}$. For this we are going to consider $y={{x}^{2}}-2x+3$ and convert it into ${{x}^{2}}=y-3+2x$. This gives 4a = 1. After this, by using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ we get $\begin{aligned} & x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)} \\\ & \Rightarrow x=\dfrac{2\pm \sqrt{4-12}}{2} \\\ \end{aligned}$ Since, the discriminant 4 – 12 is negative therefore, we will get no real x-intercepts. Here, the value of x can be determined by the formula $x=\dfrac{-b}{2a}$ which results into, $\begin{aligned} & x=\dfrac{-\left( -2 \right)}{2\left( 1 \right)} \\\ & \Rightarrow x=\dfrac{2}{2}=1 \\\ \end{aligned}$ By substituting x = 1 into $y={{x}^{2}}-2x+3$ we get $\begin{aligned} & y={{\left( 1 \right)}^{2}}-2\left( 1 \right)+3 \\\ & \Rightarrow y=1-2+3=2 \\\ \end{aligned}$ So, the ordered pair will be (1, 2) and by (i) we name it as a minimum ordered pair for given parabola. For other points we will use the process of substitution. In this process, we substitute points in place of x and get the value of y. Let us start the substitution with x = 0. So, now we will put this value in the given equation $y={{x}^{2}}-2x+3$. Thus, we will get $\begin{aligned} & y={{x}^{2}}-2x+3 \\\ & \Rightarrow y={{\left( 0 \right)}^{2}}-2\left( 0 \right)+3 \\\ & \Rightarrow y=3 \\\ \end{aligned}$ Therefore, we get one point which is (0, 3). Similarly, we take other points such as x = - 1, 1, - 2, 2 and so on. And we will collect these points as follows. Take x = - 1. So, we have $\begin{aligned} & y={{x}^{2}}-2x+3 \\\ & \Rightarrow y={{\left( -1 \right)}^{2}}-2\left( -1 \right)+3 \\\ & \Rightarrow y=1+2+3=6 \\\ \end{aligned}$ Thus, we have got (- 1, 6) as another point. Similarly, we have (1, 2) and so on. Now, after doing this process, the next step that we will follow is the process of plotting. In this process we will plot these points on the graph and we will join those points together to get the following trace.  Mostly, we do not highlight points in the graph. So, the following graph will be the right one.  **Note:** Since we have used the substitution method here, it needs concentration because we get any one point as a wrong point which will result in a wrong graph. Actually, it is not enough to collect all ordered pairs in this question as it will take time therefore, we can also use a different method. It is a direct method. In this case, the equation given to us is $y={{x}^{2}}-2x+3$. This relates to the equation of parabola $y={{x}^{2}}$. Since, this equation is positive, so the parabola will be facing upwards rather than downwards. Now, we only need to find out a few points and we are good at plotting the graph properly. It is not necessary to highlight every point on the graph as it is not possible. So, we will simply plot these points on the graph and join them together to get the required graph. Moreover, we will not find two points by using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This is because we will get a negative term inside the root of $\sqrt{{{b}^{2}}-4ac}$ as shown below. $\begin{aligned} & x=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)} \\\ & \Rightarrow x=\dfrac{2\pm \sqrt{4-12}}{2} \\\ \end{aligned}$