How do you graph the function #y=-1/4x^2+3# and identify the domain and range?

Question

Eva Adamson · Accepted Answer

See below.

The y axis intercept, or the value of the function at x = 0, is one of the important graphing points. Start with #y = -1/4x^2 + 3#. y = 3 when x = 0. The y axis is the axis of symmetry since there isn't a #x# term. This is how the parabola will appear because the coefficient of #x^2# is negative: #nnn# Find the roots of #-1/4x^2 + 3 = 0# next. #-x^2 + 12 = 0# x = +-sqrt(12) = +-2sqrt(3)# when #x^2 = 12. #( - 2sqrt(3), 0)# and #( 2sqrt(3), 0 )# are the roots. Since there are no restrictions on #x# due to the continuous nature of the function, the domain is: #{x in RR}# Because the coefficient of #x^2# is negative, the maximum value occurs when #x = 0#. This is 3. as #x -> +-oo# -1/4x^2 + 3 -> -oo## Range is therefore #{y within RR|-oo < y <= 3}# View the graph: graph{-1/4x^2 + 3 [-10, 10, 5, 5, 10]}

Ethan Adkins · Answer

undefined To graph the function $y = -\frac{1}{4}x^2 + 3$, you can start by plotting points. Choose a few x-values, calculate the corresponding y-values using the function, and then plot the points on a coordinate plane.

For the domain, it includes all real numbers, as there are no restrictions on the possible values of $x$.

For the range, notice that the coefficient of $x^2$ is negative, indicating that the parabola opens downwards. The highest point of the parabola occurs at the vertex. Since the vertex of this parabola is at $y = 3$, the range is $y \leq 3$.

After plotting the points and observing the shape of the graph, you can confirm that the graph is a downward-opening parabola with its vertex at the point (0, 3).