Question

How do you tell whether the graph opens up or down, find the vertex and find the axis of symmetry given $y=\dfrac{1}{2}{{x}^{2}}+x-2$ ?

Answer 1

Hint : We equate the given equation of parabolic curve with the general equation of ${{\left( x-\alpha \right)}^{2}}=4a\left( y-\beta \right)$ . We find the number of x intercepts and the value of the y intercept. We also find the coordinates of the focus to place the curve in the graph.

Complete step by step solution:
The given equation $y=\dfrac{1}{2}{{x}^{2}}+x-2$ is a parabolic curve.
We get $2y={{x}^{2}}+2x-4$
We convert the equation into a square form and get
$\begin{aligned} & 2y={{x}^{2}}+2x-4 \\\ & \Rightarrow {{x}^{2}}+2x+1=2y+5 \\\ & \Rightarrow {{\left( x+1 \right)}^{2}}=2\left( y+\dfrac{5}{2} \right) \\\ \end{aligned}$
We equate ${{\left( x+1 \right)}^{2}}=2\left( y+\dfrac{5}{2} \right)$ with the general equation of parabola ${{\left( x-\alpha \right)}^{2}}=4a\left( y-\beta \right)$ .
For the general equation $\left( \alpha ,\beta \right)$ is the vertex. 4a is the length of the latus rectum. The coordinate of the focus is $\left( \alpha ,\beta +a \right)$ .
This gives the vertex as $\left( -1,-\dfrac{5}{2} \right)$ . The length of the latus rectum is $4a=2$ which gives $a=\dfrac{1}{2}$ .

We have to find the possible number of x intercepts and the value of the y intercept. The curve cuts the X and Y axis at certain points and those are the intercepts.
We first find the Y-axis intercepts. In that case for the Y-axis, we have to take the coordinate values of x as 0. Putting the value of $x=0$ in the equation $2y={{x}^{2}}+2x-4$ , we get
$\begin{aligned} & 2y={{0}^{2}}+2\times 0-4=-4 \\\ & \Rightarrow y=-2 \\\ \end{aligned}$
So, the intercept point for the Y-axis is $\left( 0,-2 \right)$ . There is only one intercept on both Y-axis.
We first find the X-axis intercepts. In that case for X-axis, we have to take the coordinate values of y as 0. Putting the value of $y=0$ in the equation $2y={{x}^{2}}+2x-4$ , we get

& 2y={{x}^{2}}+2x-4 \\\ & \Rightarrow 0={{x}^{2}}+2x-4 \\\ & \Rightarrow {{\left( x+1 \right)}^{2}}=5 \\\ & \Rightarrow x=-1\pm \sqrt{5} \\\ \end{aligned}$$ The intercept point for X-axis is $ \left( -1\pm \sqrt{5},0 \right) $ . There are two intercepts for the Y-axis. The graph opens up. **The axis of symmetry for $ y=\dfrac{1}{2}{{x}^{2}}+x-2 $ is $ x=-1 $.** **Note** : The minimum point of the function $ y=\dfrac{1}{2}{{x}^{2}}+x-2 $ is $ \left( -1,-\dfrac{5}{2} \right) $ . The graph is bounded at that point. But on the other side the curve is open and not bounded. The general case of a parabolic curve is to be bounded at one side to mark the vertex.