Question

Find the vertex and intercepts of the parabola. y = x² + 4x + 5

Answer 1

Vertex V = minimum point

is found by x = -b/2a

locate a,b in equation

a= 1 , b= 4

then x = -4/2•1 = -4/2= -2

Now find vertex y = (4ac - b^2)/ 4a^2

. y = ( 4•1•5 - 16)/ 4•1

. y= (20 - 16) / 4

. y = 4/4 = 1

Then Vertex is located at (-2,1)

Now find intercepts x,y

For y- intercept , x= 0, then y = 0+0 + 5 = 5

For x-intercept ,y = 0, then x² + 4x + 5 = 0

BUT , this equation have no x- intercept, because never touches y= 0