# How do you find the axis of symmetry, and the maximum or minimum value of the function y = x^2 + 3x ?

Jan 18, 2017

Of form that has a minimum duo to the coefficient of ${x}^{2}$ being positive.

$\implies {P}_{\text{minimum}} \to \left(x , y\right) = \left(- \frac{3}{2} , - \frac{9}{4}\right)$

Axis of symmetry is: $x = - \frac{3}{2}$

#### Explanation:

The ${x}^{2}$ is positive so the general shape of the graph is $\cup$ thus we have a minimum.

Just for a moment, suppose we had $- {x}^{2}$ in that scenario the graph would be of form $\cap$ and thus a maximum.

Consider the general case of $y = a {x}^{2} + b x + c$

Write this as: $a \left({x}^{2} + \frac{b}{a} x\right) + c$

The axis of symmetry is at $x = \left(- \frac{1}{2}\right) \times \left(\frac{b}{a}\right)$

So in this case we have:

${x}_{\text{symmetry") = (-1/2)xx3/1 = -3/2 = x_("minimum}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Set x =-3/2" "=>" "y_("minimum")=(-3/2)^2+3(-3/2) = -2 1/4

$\implies {P}_{\text{minimum}} \to \left(x , y\right) = \left(- \frac{3}{2} , - \frac{9}{4}\right)$

Jan 18, 2017

Axis of symmetry $x = - \frac{3}{2}$
Minimum value $- \frac{9}{4}$
Maximum value $\infty$

#### Explanation:

To put the equation in standard form, x-terms are made into a perfect square. In the given expression consider the coefficient of x, which is 3. Half of this is $\frac{3}{2}$. Square this to have $\frac{9}{4}$ Now add and subtract $\frac{9}{4}$ on the right side of the expression, as shown below

y= x^2 +3x +9/4 -9/4 =(x+3/2)^2 -9/4

The equation in this form represents a vertical parabola, opening up, with vertex at $\left(- \frac{3}{2} , - \frac{9}{4}\right)$

The axis of symmetry is $x = - \frac{3}{2}$

Minimum value is $- \frac{9}{4}$, maximum value is $\infty$