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

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

x=-5

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 = - b 2a for the equation of the form y=a x2 + bx+c.

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

x = - 10 2⋅1 = - 5

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