Question

How do you graph the parabola $y=2{{\left( x-1 \right)}^{2}}-5$ using vertex, intercepts and additional points.

Answer 1

In the given equation of the parabola $y=2{{\left( x-1 \right)}^{2}}-5$, we have to plot the graph using vertex points, intercepts and additional points. We know that the general vertex form of the quadratic equation of the parabola, comparing the general vertex form to the given equation of the parabola, we can get the value of h and k, which is the vertex of the parabola. We also know that at x-intercept y is 0 and vice versa. After finding the points, we can plot those points in the graph.

Complete step by step answer:
We know that the general vertex form of quadratic equation of parabola is,
$y=a{{\left( x-h \right)}^{2}}+k$ ……… (1)
Where, a is the coefficient of square of x, h is the axis of symmetry and k is the minimum/maximum value of the function.
We also know that the given equation of the parabola is,
$y=2{{\left( x-1 \right)}^{2}}-5$ ……… (2)
Now we can compare equation (1) and (2), we get
The value of a = 2, the value of h = 1 and the value of k = -5.
We know that the vertex of the parabola is the point where the curve turns, which is always at maximum or minimum value.
We also know that the coordinate of the vertex is $\left( h,k \right)$, where we already know the values.
Therefore, the vertex is $\left( 1,-5 \right)$.
Now we can find the intercept value,
We know that for x-intercept, y=0 from (2), we get

& \Rightarrow 0=2{{\left( x-1 \right)}^{2}}-5 \\\ & \Rightarrow \dfrac{5}{2}={{\left( x-1 \right)}^{2}} \\\ \end{aligned}$$ We can square on both sides, we get $$\begin{aligned} & \Rightarrow \left( x-1 \right)=\pm \sqrt{\dfrac{5}{2}} \\\ & \Rightarrow x=1\pm \sqrt{\dfrac{5}{2}} \\\ \end{aligned}$$ Therefore, the x coordinates are $$\left( 1+\sqrt{\dfrac{5}{2}},0 \right),\left( 1-\sqrt{\dfrac{5}{2}},0 \right)$$. We can write in decimal form, we get x-coordinates $$\left( 2.5811,0 \right),\left( -0.5811,0 \right)$$ We also know that for y-intercept, x=0 from (2), we get $$\begin{aligned} & \Rightarrow y=2{{\left( 0-1 \right)}^{2}}-5 \\\ & \Rightarrow y=-3 \\\ \end{aligned}$$ Therefore, the y-coordinate is $$\left( 0,-3 \right)$$ Now we can find other points. We know that to find other points, we can choose value for x and y, For x = -1, $$\begin{aligned} & \Rightarrow y=2{{\left( -1-1 \right)}^{2}}-5 \\\ & \Rightarrow y=3 \\\ \end{aligned}$$ For x = 3, $$\begin{aligned} & \Rightarrow y=2{{\left( 3-1 \right)}^{2}}-5 \\\ & \Rightarrow y=3 \\\ \end{aligned}$$ Therefore, the other points are $$\left( -1,3 \right)\left( 3,3 \right)$$ Now we can plot the vertex $$\left( 1,-5 \right)$$, x-coordinates $$\left( 2.5811,0 \right),\left( -0.5811,0 \right)$$, y-coordinate $$\left( 0,-3 \right)$$ and other points $$\left( -1,3 \right)\left( 3,3 \right)$$.  **Note:** Student make mistakes in finding the vertex form $${{\left( x-h \right)}^{2}}$$, here in this question we have the same form, if we have positive sigh over there, the we should know that the value of h will be negative. We know that to find x-intercept the value of y is 0 and vice versa.