Respuesta :
Answer and Explanation:
Given : The quadratic function [tex]f(x)=-x^2+x+12[/tex]
To find : Determine the following ?
Solution :
The x -intercept are where f(x)=0,
So, [tex]-x^2+x+12=0[/tex]
Applying middle term split,
[tex]-x^2+4x-3x+12=0[/tex]
[tex]-x(x-4)-3(x-4)=0[/tex]
[tex](x-4)(-x-3)=0[/tex]
[tex]x=4,-3[/tex]
The x-intercepts are (4,0) and (-3,0).
The smallest x-intercept is x=-3
The largest x-intercept is x=4
The y -intercept are where x=0,
So, [tex]f(0)=-(0)^2+0+12[/tex]
[tex]f(0)=12[/tex]
The y-intercept is y=12.
The quadratic function is in the form, [tex]y=ax^2+bx+c[/tex]
On comparing, a=-1 , b=1 and c=1 2
The vertex of the graph is denote by (h,k) and the formula to find the vertex is
For h, The x-coordinate of the vertex is given by,
[tex]h=-\frac{b}{2a}[/tex]
[tex]h=-\frac{1}{2(-1)}[/tex]
[tex]h=\frac{1}{2}[/tex]
For k, The y-coordinate of the vertex is given by,
[tex]k=f(h)[/tex]
[tex]k=-h^2+h+12[/tex]
[tex]k=-(\frac{1}{2})^2+\frac{1}{2}+12[/tex]
[tex]k=-\frac{1}{4}+\frac{1}{2}+12[/tex]
[tex]k=\frac{-1+2+48}{4}[/tex]
[tex]k=\frac{49}{4}[/tex]
The vertex of the function is [tex](h,k)=(\frac{1}{2},\frac{49}{4})[/tex]
The x-coordinate of the vertex i.e. [tex]x=-\frac{b}{2a}[/tex] is the axis of symmetry,
So, [tex]x=-\frac{b}{2a}=\frac{1}{2}[/tex] (solved above)
The axis of symmetry is [tex]x=\frac{1}{2}[/tex].
