Question

Question #576da

Answer 1

`x=-6,x=-4,x=-5,(-5,-1)`

Explanation:

`"to find the x-intercepts solve "x^2+10x+24=0`

`"the factors of 24 which sum to + 10 are + 4 and + 6"`

`rArr(x+6)(x+4)=0`

`"equate each factor to zero and solve for x"`

`x+6=0rArrx=-6larrcolor(red)" x-intercept"`

`3x+4=0rArrx=-4larrcolor(red)" x-intercept"`

`"given the parabola in standard form "ax^2+bx+c`

`"the x-coordinate of the vertex/ axis of symmetry is"`

`x_(color(red)"vertex")=-b/(2a)`

`y=x^2+10x+24" is in standard form"`

`"with "a=1,b=10,c=24`

`x_(color(red)"vertex")=-10/2=-5`

`"substitute this value into the equation for y"`

`rArry_(color(red)"vertex")=(-5)^2+10(-5)+24=-1`

`rArrcolor(magenta)"vertex "=(-5,-1)`

`"axis of symmetry is "x=-5`
graph{(y-x^2-10x-24)(y-1000x-5000)=0 [-10, 10, -5, 5]}

Answer 2

The x intercepts are `x_1=-6` and `x_2=-4`
The axis of symmetry is: `x=-5`
The vertex is `V=(-5;-1)`

Explanation:

In order to find the x intercepts, we will substitute `y=0` and then solve the equation `x^2+10x+24=0`.
We will use the short quadratic formula `x=-b/2+-sqrt((b/2)^2-c`, useful when b is even and a=1 in the general form `ax^2+bx+c=0`

Then `x=-5+-sqrt(5^2-24)=-5+-1`

and the x intercepts are `x_1=-6` and `x_2=-4`

The axis of symmetry is:

`x=-b/(2a)=-10/2=-5`

We simply will get the second coordinate of the vertex by substituting x=-5 in the given F(x):

Then `V=(-5;F(-5))=(-5;25-50+24)=(-5;-1)`
graph{x^2+10x+24 [-10, 5, -5, 5]}