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

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Answer 1 · 2017-01-15T04:04:51

Refer the explanation section

Given -

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

Find the vertex first-

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

$x = \frac{- b}{2 a}$ $x=(-b)/(2a)$

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

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

$x = \frac{- (0)}{2 \times (- 1)} = 0$ $x=(-(0))/(2 xx (-1))=0$

At $x = 0$ $x=0$

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

The vertex is $(0, 3)$ $(0, 3)$

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

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

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 $y$ $y$ value.

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

See the Table

See the graph

How do you find the important parts of the equation to graph the function y=−x2+3y = -x^2 + 3?