# How do you find the important parts of the equation to graph the function y = -x^2 + 3?

Jan 15, 2017

Refer the explanation section

#### Explanation:

Given -

$y = - {x}^{2} + 3$

Find the vertex first-

If the quadratic equation is in the form $y = a {x}^{2} + b x + c$, then it is given by

$x = \frac{- b}{2 a}$

Since there is no $b x$ term in the given equation, we shall supply it

$y = - {x}^{2} + 0 x + 3$

$x = \frac{- \left(0\right)}{2 \times \left(- 1\right)} = 0$

At $x = 0$

$y = - {0}^{2} + 3 = 3$

The vertex is $\left(0 , 3\right)$

Then we have to decide whether the curve is concave downwards or upwards. That is given by the second derivative.

$\frac{\mathrm{dy}}{d} = - 2 x$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - 2 < 0$

Since the second derivative is less than zero, the curve is concave downwards.

Take two points to the right of zero [ x-coordinate of the vertex] and two points to the left of zero. Find the corresponding $y$ value.

Plot the pairs in a graph sheet. Join all the points.  