Question

Can someone please find the line of symmetry, vertex, roots, y-intercept and minimum or maximum for y=(x+3)^2-4 I need it now

Answer 1

The equation of the parabola is given is the Vertex Form. The general form of a quadratic equation in Vertex Form is:

`y=a(x+b)^(2)+c`

So for our equation,
`a=1, b=3,` and
`c=-4`. Now let's solve the question.

1. Line of Symmetry:

Line of symmetry is given as
`x=-b`, so our line of symmetry is
`x=-3`.

2. Vertex:

Vertex is given as
`(-b,c)`, so our vertex is
`(-3,-4)`.

3. Roots:

We find the roots by setting
`y=0`. Thus, we have

`0=(x+3)^(2) -4\\4=(x+3)^(2)\\`

So,

`x+3=2` and
`x+3=-2`

So, solving these 2 equations we have
`x=-1, -5`

4. Y-Intercept:

To find y-intercept, we set
`x=0`. So we have

`y=(0+3)^(2)-4\\y=5`

5. Minimum/Maximum:

A quadratic equation has minimum if
`a` is positive and maximum is
`a` is negative. Hence, this function has a minimum since
`a` is positive. The value of the minimum is
`y=c`. So for our question, the minimum is
`y=-4`.

ANSWERS:

1. Line of Symmetry:
`x=-3`

2. Vertex:
`(-3,-4)`

3. Roots:
`x=-1, -5`

4. y-intercept:
`y=5`

5. Minimum Value:
`y=-4`