Question

How do you find the axis of symmetry, graph and find the maximum or minimum value of the function `y=x^2+4x-1`?

Answer 1

Explanation is given below.

Explanation:

`y=x^2+4x-1`

To solve such problems, go for completing the square. If not then find vertex using `h=-b/(2a)` and `k` by substituting the `x=h` in the given equation.

The `x=h` would be the axis of symmetry. If the graph opens up there would be a minimum. If the graph opens down then there would be a maximum.

The coefficient of `x^2` is positive then the graph opens up as is the case in this problem that means we have a minimum value and that value is `k` value of the vertex.

Let me show it by completion of square.

`y=x^2+4x-1`
I would first add `1` to both sides, this is done to keep `x^2+4x` on one side which we would convert to a perfect square.

`y+1=x^2+4x-cancel(1)+cancel(1)`
`y+1=x^2+4x`

Next step is to take the coefficient of `x` and divide it by `2`. Square the result and add both sides.

`y+1+(4/2)^2=x^2+4x+(4/2)^2`
`y+1+(2)^2=x^2+4x+2^2`
`y+1+4=(x+2)^2`
`y+5=(x+2)^2`
Subtract `5` both the sides we get
`y+cancel(5)-cancel(5)=(x+2)^2-5`

`y=(x+2)^2-5` Vertex form

Vertex `(-2,-5)`
Axis of symmetry `x=-2`
Minimum at `(-2,-5)` the minimum value is `-5`

Answer 2

Plot the graph using Graph tool given above.

Explanation:

graph{y=x^2+4x-1 [-20, 20, -10, 10]}
From the Graph:
Vertex is located at `(−2,−5)`
Axis of symmetry `x=−2`
Minimum at `(−2,−5)`.