Hello there!
Today we are just going to graph this quadratic function.
First let's find the vertex. There are two ways to find the vertex, but we will use the formula -b/2a today!
f(x)=x^2+10x+24
plug this into -b/2a where a=1 and b=10. (I found out what a and b equal by looking at the each term's coefficient.)
-10/2(1)
-10/2
-5
So the x value for our vertex is -5, so now to find the y-value, we need to plug in -5 in place of x in the equation...
f(x)=x^2+10x+24
f(-5)=(-5)^2+10(-5)+24
f(-5)=25-50+24
f(-5)=-25+24
f(-5)=-1
So the vertex is at (-5,-1)
Look at the table...
X] -5
Y] -1
Now we need to plug in number around the vertex to the equation.
X] -3 -4 -5 -6 -7
Y] -1
f(x)=x^2+10x+24
f(-4)=(-4)^2+10(-4)+24
f(-4)=16-40+24
f(-4)=-24+24
f(-4)=0
f(-3)=(-3)^2+10(-3)+24
f(-4)=9-30+24
f(-4)=-21+24
f(-4)=3
X] -3 -4 -5 -6 -7
Y] 3 0 -1
Now because the parabola is symmetrical, we don't have to plug in the other points to the equation. Watch what I do...
X] -3 -4 -5 -6 -7
Y] 3 0 -1 0 3
I really hope this helps!
Best wishes :)
Today we are just going to graph this quadratic function.
First let's find the vertex. There are two ways to find the vertex, but we will use the formula -b/2a today!
f(x)=x^2+10x+24
plug this into -b/2a where a=1 and b=10. (I found out what a and b equal by looking at the each term's coefficient.)
-10/2(1)
-10/2
-5
So the x value for our vertex is -5, so now to find the y-value, we need to plug in -5 in place of x in the equation...
f(x)=x^2+10x+24
f(-5)=(-5)^2+10(-5)+24
f(-5)=25-50+24
f(-5)=-25+24
f(-5)=-1
So the vertex is at (-5,-1)
Look at the table...
X] -5
Y] -1
Now we need to plug in number around the vertex to the equation.
X] -3 -4 -5 -6 -7
Y] -1
f(x)=x^2+10x+24
f(-4)=(-4)^2+10(-4)+24
f(-4)=16-40+24
f(-4)=-24+24
f(-4)=0
f(-3)=(-3)^2+10(-3)+24
f(-4)=9-30+24
f(-4)=-21+24
f(-4)=3
X] -3 -4 -5 -6 -7
Y] 3 0 -1
Now because the parabola is symmetrical, we don't have to plug in the other points to the equation. Watch what I do...
X] -3 -4 -5 -6 -7
Y] 3 0 -1 0 3
I really hope this helps!
Best wishes :)