What are the zeros, equation of the axis of symmetry and vertex of #y=x^2 + 5x+ 8#?

To find the zeros (also called solutions or roots) if they exist, we set #y=0# and solve for x:

#y=x^2 + 5x+ 8#

#0=x^2 + 5x+ 8#

We have to use the quadratic formula to factor:

#ax^2 + bx +c#

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#a=1#
#b=5#
#c=8#

#x = (-5+-sqrt(5^2-4*1*8))/(2*1)#

#x = (-5+-sqrt(25-32))/(2)#

#x = (-5+-sqrt(-7))/(2)#

Since the #sqrt(-7)# is an imaginary number there all no real roots or zeros.

The formula for axis of symmetry (aos):

#aos=(-b)/(2a)#

#aos=(-5)/(2*1) = -5/2#

Fromula for vertex is:

#(aos, f(aos))#, remember #y=f(x)#

#f(x)=x^2 + 5x+ 8#

#f(-5/2)=(-5/2)^2 + 5(-5/2)+ 8#

#f(-5/2)=7/4#

So the vertex #= (-5/2, 7/4)#

When given a quadratic of the form #y = ax^2+bx+c#

The equation of the axis of symmetry is:

#x = -b/(2a)#

The x-coordinate of the vertex, #h#, has the same value as the axis of symmetry:

#h = -b/(2a)#

The y-coordinate of the vertex, #k#, is the function evaluated at #h#:

#k = ak^2+bk + c#

The zeros can be found using the quadratic formula:

#x_1 = (-b-sqrt(b^2-4(a)(c)))/(2a)# and #x_2 = (-b+sqrt(b^2-4(a)(c)))/(2a)#

Given: #y=x^2 + 5x+ 8#

Please observe that #a = 1, b = 1, and c = 8#

The equation of the axis of symmetry is:

#x = -5/2#

Compute the vertex:

#h = -5/2#

#k = (-5/2)^2+5(-5/2)+8#

#k = 7/4#

The vertex is #(-5/2, 7/4)#

Compute the zeros:

#x_1 = (-5-sqrt(5^2-4(1)(8)))/(2(1))# and #x_2 = (-5+sqrt(5^2-4(1)(8)))/(2(1))#

#x_1 = -5/2-(sqrt7i)/2# and #x_2 = -5/2+(sqrt7i)/2#