Chapter 7, Problem 1RE

To determine

To sketch: Graph of the given equation.

Answer to Problem 1RE

Solution:

Explanation of Solution

Explanation: Given: The equation $-x^2=y$ .

Concept Used:

1. For the general equation of parabola $x^2=4py$ , the parabola opens downward when $p<0$ and opens upward when $p>0$ .
2. Vertex of the parabola of the form $x^2=4py$ is $(0,0)$ , and axis of symmetry is $x=0$ .
3. Get two point on the parabola on each side of the axis of symmetry.

Calculation: Comparing the given equation to general equation of parabola $x^2=4py$ , we get,

  $\begin{array}{l}-x^2=y\\ x^2=-y=4py\\ 4p=-1\\ p=-\frac{1}{4}\end{array}$

We know that the parabola opens downward when $p<0$ and here, we have $p=-\frac{1}{4}$ . So, the parabola opens downward.

The vertex of the parabola is at $(0,0)$ , and axis of symmetry is $x=0$ , that means, the axis of symmetry is the $y-\text{axis}$ .

To find two points on the parabola, let us choose a random value for $y$ as $y=-1$ and substitute in the given equation and solve for $x$ .

  $\begin{array}{l}-x^2=-1\\ \Rightarrow \frac{-x^2}{-1}=\frac{-1}{-1}\left(\begin{array}{l}\text{To get rid of the negative, we divide both }\\ \text{sides of the equation by -1}\text{.}\end{array}\right)\\ \Rightarrow x^2=1\\ \Rightarrow x=\pm \sqrt{1}\\ \Rightarrow x=\pm 1\end{array}$

Therefore, we get the two points as $\left(-1,-1\right)$ , and $\left(1,-1\right)$ .

Thus, joining the points $\left(-1,-1\right)$ , $\left(1,-1\right)$ and the vertex $(0,0)$ , we can draw the given equation as shown below.

Conclusion: The graph of the given equation is a downward parabola with vertex at $\left(0,0\right)$ .