Question

y=x^{2} +4x-7 determine the axis of symmetry the vertex point and find the roots useing the QUADRATIC FORMULA

Answer 1

Axis of symmetry is x = -2

(-2, -11) is the vertex

Roots are

-2 +
`√(11)`, -2 -
`√(11)`

Explanation:

a) Axis of symmetry equation is: x = -b/2a

Original equation: y =
`x^2+4x-7`

So...

x= -4/2(1) = -2

Axis of symmetry is x = -2

b) Vertex Point can be found by plugging into original equation.

`y =x^2+4x-7`

=> y =
`(-2)^2+4(-2)-7`

=> 4 - 8 - 7 = y

=> -11 = y

(-2, -11) is the vertex

c) quadratic formula is
`x = (-b +-√(b^2-4ac) )/(2a)`

Lets plug in the values.

`\frac{-4 +- \sqrt{4^(2)-4(1)(-7) } }{2(1)}`

`(-4 +- √(16+28 ) )/(2)`

`(-4 +- √(44 ) )/(2)`

`√(44) = 2√(11)`

-4 +- 2
`√(11)`

--------------

2

Here, i will factor out 2, and take out the +- and make it into two equations

2(-2 +
`√(11)`) / 2 = -2 +
`√(11)`

2(-2 -
`√(11)`) / 2 = -2 -
`√(11)`

x= -2 +
`√(11)`

or

x = -2 -
`√(11)`

Roots are

-2 +
`√(11)`, -2 -
`√(11)`

Answer 2

Explanation:

y=x²+4x-7

=x²+4x+4-4-7

=(x+2)²-11

axis of symmetry is x=-2

vertex is (-2,-11)

x²+4x-7=0

`x=(-4 \pm√(4^2-4(1)(-7)) )/(2(1)) \\=(-4 \pm √(16+28) )/(2) \\=(-4 \pm 2√(11) )/(2) \\=-2 \pm√(11)`