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

Feb 11, 2016

Explanation is given below.

#### Explanation:

$y = {x}^{2} + 4 x - 1$

To solve such problems, go for completing the square. If not then find vertex using $h = - \frac{b}{2 a}$ 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.

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

$y + 1 = {x}^{2} + 4 x - \cancel{1} + \cancel{1}$
$y + 1 = {x}^{2} + 4 x$

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

$y + 1 + {\left(\frac{4}{2}\right)}^{2} = {x}^{2} + 4 x + {\left(\frac{4}{2}\right)}^{2}$
$y + 1 + {\left(2\right)}^{2} = {x}^{2} + 4 x + {2}^{2}$
$y + 1 + 4 = {\left(x + 2\right)}^{2}$
$y + 5 = {\left(x + 2\right)}^{2}$
Subtract $5$ both the sides we get
$y + \cancel{5} - \cancel{5} = {\left(x + 2\right)}^{2} - 5$

$y = {\left(x + 2\right)}^{2} - 5$ Vertex form

Vertex $\left(- 2 , - 5\right)$
Axis of symmetry $x = - 2$
Minimum at $\left(- 2 , - 5\right)$ the minimum value is $- 5$

Feb 11, 2016

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).