1)
We are given a equation p(x) which is given as:
[tex]p(x)=30x+5x^2[/tex]
Sandra's function is a quadratic function whose graph is a upward open parabola with vertex at (-3,-45).
2)
We know that the vertex of a equation of the type:
[tex]y-k=a(x+h)^2[/tex]
is (h,k)
We have the equation as:
[tex]p(x)=5(x+3)^2-45\\\\p(x)-(-45)=5(x-(-3))^2[/tex]
Hence, the vertex is:
(-3,-45).
3)
As the parabola is a upward open parabola hence, the parabola has a minimum value but not the maximum values as the function is unbounded above.
Hence, the vertex is a minimum.
4)
The axis of symmetry of a parabolic function is a line which act as a mirror for the parabola.
We know that for the general equation of the type:
[tex]y=ax^2+bx+c[/tex]
The axis of symmetry is:
[tex]x=\dfrac{-b}{2a}[/tex]
We have a=5 and b=30
Hence, the axis of symmetry is:
[tex]x=-\dfrac{30}{2\times 5}\\\\\\x=-3[/tex]
Hence, axis of symmetry is:
x= -3
The vertex of the parabola are (-3, -45).
In the attached graph the vertex of the parabola and parabola opens upward then the parabola has the maximum value.
The axis of symmetry is -3.
The standard vertex form of the equation is;
[tex]\rm y=a(x-h)^2+k[/tex]
Where h and k are the vertexes of the equation.
Given information
Sandra wrote p(x) = 30x + 5x2 in vertex form.
1. These are the following steps to convert the equation into vertex form;
[tex]\rm P(x)=30x+5x^2\\\\P(x)=5x^2+30x+45-45\\\\ P(x)=5(x^2+6x+9)-45\\\\P(x)=5(x+3)^2-45[/tex]
On comparing with standard vertex equation (-3, -45) are the vertex of the equation.
2. In the attached graph the vertex of parabola and parabola opens upward then the parabola has the maximum value.
3. The axis of symmetry of this function is;
[tex]\rm Axis \ of \ symmetry = \dfrac{-b}{2a}\\\\ Axis \ of \ symmetry =\dfrac{-30}{5(2)}\\\\ Axis \ of \ symmetry =\dfrac{-30}{10}\\\\ Axis \ of \ symmetry =-3[/tex]
The axis of symmetry is -3.
To know more about vertex form click the link given below.
https://brainly.com/question/1587077
