How do you find the axis of symmetry, graph and find the maximum or minimum value of the function #f(x)= 3x^2#?

Question

Hazel Anglin · Accepted Answer

Refer to the explanation.

#f(x)=3x^2# is a quadratic equation in standard form: #y=ax^2+bx+c#, where: #a=3#, #b=0#, and #c=0# Axis of symmetry: the vertical line that divides the parabola into two equal halves. For a quadratic equation in standard form, the formula for the axis of symmetry is: #x=(-b)/(2a)# Plug in the known values. #x=(0)/(2*3)# Simplify. Axis of symmetry: #x=0# This means that the axis of symmetry is on the x-axis, where #x=0#. This is also the #x#-value of the vertex. Vertex: the maximum or minimum point on the parabola. To determine the #y#-value of the vertex, substitute #0# for #x# and solve for #y#. #y=3x^2# Plug in #0# for #x#. #y=3(0)^2=0# Vertex: #(0,0)# Since the vertex does not cross the x-axis, we don't have x-intercepts in terms of #(x,0)#, but we can determine other points of #(x,y)# on the parabola. I will propose six values for #x# (3 positive and 3 negative), and plug them into the equation and solve for #y#. This will give six symmetrical points on the parabola. #x=0.5# #y=3(0.5)^2=0.75# Point 1: #(0.5,0.75)# #x=-0.5# #y=3(-0.5)^2=0.75# Point 2: #(-0.5,0.75)# #x=1# #y=3(1)^2=3# Point 3: #(1,3)# #x=-1# #y=3(-1)^2# #y=3# Point 4: #(-1,3)# #x=2# #y=3(2)^2# #y=12# Point 5: #(2,12)# #x=-2# #y=3(-2)^2# #y=12# Point 6: #(-2,12)# Plot the vertex and the other six points. Sketch a parabola through the points. Do not connect the dots. graph{y=3x^2 [-10, 10, -5, 5]}

Abigail Allen · Answer

undefined The axis of symmetry for a quadratic function in the form $f(x) = ax^2 + bx + c$ is given by the equation $x = -\frac{b}{2a}$.

For the function $f(x) = 3x^2$, $a = 3$ and $b = 0$.

So, the axis of symmetry is $x = -\frac{0}{2 \times 3} = 0$.

To graph the function, you plot points for different values of $x$ and calculate the corresponding $y$ values using the function $f(x) = 3x^2$.

The function $f(x) = 3x^2$ represents a parabola that opens upwards because the coefficient of $x^2$ is positive.

Since the coefficient of $x^2$ is positive, the vertex of the parabola is the minimum point. The minimum value of the function occurs at the vertex.

For the function $f(x) = 3x^2$, the vertex is at $x = 0$ (axis of symmetry) and $y = f(0) = 3(0)^2 = 0$.

So, the minimum value of the function $f(x) = 3x^2$ is $0$, and it occurs at $x = 0$.