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

1 Answer
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=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 TableSee the Table

See the graphSee the graph