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