How do you graph #y=-x^2+4#?

1 Answer
marfre
Apr 12, 2017

Point plotting or finding the vertex and axis of symmetry

Explanation:

Point plotting:
Create a table of #x# and #y# values.

Since #x# is the independent variable in the equation (#y# depends on the #x# variables selected), you can select any #x# value and find the corresponding #y# value using the equation #y = -x^2 + 4#

#"x|"-3"|"-2"|"-1"| "0"| "1"| "2"| "3"|"#
#"y|"-5"| "0"| "3"| "4"| "3"| "0"|"-5"|"#

Plot these points on a coordinate plane and connect the points with an arc.

Finding the vertex and axis of symmetry:

A parabola can be graphed easily when it is in the standard form/vertex form #y = a(x - h)^2 +k# where the vertex #= (h, k)# and the axis of symmetry is #x = h#

First put the equation in general form #Ax^2 +Bx + C = 0#

A negative #A# value means the parabola opens downward, a positive #A# value means the parabola opens upward.

#h = -B/(2A)#

For #y = -x^2 + 4," " A = -1, B = 0, C = 4; " " h = 0/-2 = 0#

So the axis of symmetry is #x = 0#

#k = f(h) = f(0) = -(0)^2 + 4 = 4#

So the vertex is #(0, 4)#

graph{-x^2+4 [-12.66, 12.66, -6.33, 6.33]}

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