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

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Answer 1 · 2017-01-15T04:04:51

Refer the explanation section

Given -

$y=-x^2+3$

Find the vertex first-

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

$x=(-b)/(2a)$

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

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

$x=(-(0))/(2 xx (-1))=0$

At $x=0$

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

The vertex is $(0, 3)$

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

$dy/d=-2x$
$(d^2y)/(dx^2)=-2<0$

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 $y$ 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 = -x^2 + 3y = -x^2 + 3?