Question

Consider the quadratic function f(x) = -x^2 + 1x + 12. Determine the following: (enter all numerical answers as integers, fractions, or decimals): The smallest -intercept is 2 = The largest x-intercept is z = The y-intercept is y = The vertex is ( The line of symmetry has the equation Preview

Answer 1

Answer and Explanation:

Given : The quadratic function $f(x)=-x^2+x+12$

To find : Determine the following ?

Solution :

The x -intercept are where f(x)=0,

So, $-x^2+x+12=0$

Applying middle term split,

$-x^2+4x-3x+12=0$

$-x(x-4)-3(x-4)=0$

$(x-4)(-x-3)=0$

$x=4,-3$

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, $f(0)=-(0)^2+0+12$

$f(0)=12$

The y-intercept is y=12.

The quadratic function is in the form, $y=ax^2+bx+c$

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,

$h=-\frac{b}{2a}$

$h=-\frac{1}{2(-1)}$

$h=\frac{1}{2}$

For k, The y-coordinate of the vertex is given by,

$k=f(h)$

$k=-h^2+h+12$

$k=-(\frac{1}{2})^2+\frac{1}{2}+12$

$k=-\frac{1}{4}+\frac{1}{2}+12$

$k=\frac{-1+2+48}{4}$

$k=\frac{49}{4}$

The vertex of the function is $(h,k)=(\frac{1}{2},\frac{49}{4})$

The x-coordinate of the vertex i.e. $x=-\frac{b}{2a}$ is the axis of symmetry,

So, $x=-\frac{b}{2a}=\frac{1}{2}$ (solved above)

The axis of symmetry is $x=\frac{1}{2}$.