Question #576da

2 Answers
Aug 25, 2017

#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]}

Aug 25, 2017

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]}