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

Question

Hazel Anglin · Accepted Answer

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

Ethan Adkins · Answer

undefined To find the axis of symmetry for the quadratic equation y = x^2 + 5x + 3, use the formula: x = -b / (2a), where a is the coefficient of the x^2 term and b is the coefficient of the x term. In this case, a = 1 and b = 5, so the axis of symmetry is x = -5 / (2*1) = -5/2.

To find the vertex, substitute the x-coordinate of the axis of symmetry into the equation to find the corresponding y-coordinate. In this case, when x = -5/2:
y = (-5/2)^2 + 5*(-5/2) + 3
y = 6.25 - 12.5 + 3
y = -3.25

So, the vertex is (-5/2, -3.25).

To find the x-intercepts, set y = 0 and solve for x. In this case:
0 = x^2 + 5x + 3

Use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a). Substitute a = 1, b = 5, and c = 3 into the formula:
x = (-5 ± √(5^2 - 4*1*3)) / (2*1)
x = (-5 ± √(25 - 12)) / 2
x = (-5 ± √13) / 2

So, the x-intercepts are (-5 + √13)/2 and (-5 - √13)/2.