What is the line of symmetry for the parabola whose equation is y = x2 + 10x + 25?

The line of symmetry passes through the vertex of the parabola. The x coordinate of the vertex is found with the formula x=−b2a for the equation of the form y=ax2+bx+c.

In this example, a=1, b=10 and c=25

x=−102⋅1=−5

Because the line of symmetry passes through the vertex, the equation of the line is x=−5.

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apologiabiology apologiabiology · Answer 2 · 2017-05-05T19:48:11+02:00

the line for axis of symmetry can be found with 2 ways
1 long way (finding x value of vertex)
2. short way (finding axis of symmetry using equation)

1. long way
complete the square
y=x^2+10x+25
take 1/2 of lienar coefnent and square it
10/2=5, 5^2=25
intersting, we have 25 already
factor perfect square
y=(x+5)^2
in form
y=a(x-h)^2+k
vertex is (h,k)
axis of symmetry is x=h
so
y=1(x-(-5))^2+0
axis of symmetry is x=-5

short way
for
y=ax^2+bx+c
the x value of the vertex or the axis of symmetry is -b/(2a)

given
y=1x^2+10x+25
a=1
b=10

-b/(2a)=-10/(2*1)=-10/2=-5
the axis of symmety is x=-5