Question

How do you find the axis of symmetry, vertex and x intercepts for `y=x^2+5x+3`?

Answer 1

Vertex is at `(-2.5 , -3.5)` , Axis of symmetry is `x = -2.5`
x intercepts are at `(-4.3,0) & ( -0.7,0 )`

Explanation:

`y=x^2+5x+3 ; a= 1 , b=5 , c=3 :. -b/(2a)= -5/2= -2.5`
x cordinate of vertex is `-b/(2a) =-2.5`
y cordinate of vertex is `y= (-2.5)^2+ 5*(-2.5) +3 = -3.25`
Vertex is at `(-2.5 , -3.5)`
Axis of symmetry is `x = -2.5`
x intercepts is obtained by putting `y=0` in the eqation.
`x^2+5x+3=0 ; x = (-b +- sqrt (b^2-4ac))/(2a)= (-5 +- sqrt (25-12))/2= -5/2+- sqrt 13/2 :. x ~~ -4.303 , x ~~ -0.697` graph{x^2+5x+3 [-10, 10, -5, 5]} [Ans]