# How does the graph of y=ax^2+bx+c representing a parabola depend on a, the coefficient of x^2?

Feb 2, 2017

#### Explanation:

A quadratic equation of type $y = a {x}^{2} + b x + c$, which is equation of a parabola, can be written in vertex form i.e.

$y = a {\left(x - h\right)}^{2} + k$

For example, let us have the equation $y = - 3 {x}^{2} - 6 x + 9$, them we can write it as

$y = - 3 \left({x}^{2} + 2 x + 1\right) + 3 + 9 = - 3 {\left(x + 1\right)}^{2} + 12$

Similarly $y = 2 {x}^{2} + 3 x - 7 = 2 \left({x}^{2} + \frac{3}{2} x\right) - 7$

= $2 \left({x}^{2} + 2 \times \frac{3}{4} x + {\left(\frac{3}{4}\right)}^{2}\right) - 2 \times {\left(\frac{3}{4}\right)}^{2} - 7$

= $2 {\left(x + \frac{3}{4}\right)}^{2} - \frac{65}{8} = 2 {\left(x + \frac{3}{4}\right)}^{2} - 8 \frac{1}{8}$

Observe that ${\left(x - h\right)}^{2} \ge 0$ and hence

if $a > 0$, $a {\left(x - h\right)}^{2}$ is positive and the minimum value of $y$, when $x = h$ is $k$ i.e. the parabola rises on either side and has lowest point as $\left(h , k\right)$. For example, $y = 2 {x}^{2} + 3 x - 7 = 2 {\left(x + \frac{3}{4}\right)}^{2} - 8 \frac{1}{8}$ and graph of $y = 2 {x}^{2} + 3 x - 7$ has minima at $\left(- \frac{3}{4} , 8 \frac{1}{8}\right)$ and observe that $\left(- \frac{3}{4} , 8 \frac{1}{8}\right)$ appears as a vertex.
graph{2x^2+3x-7 [-4, 4, -11.36, 8.64]}

but if $a < 0$, $a {\left(x - h\right)}^{2}$ is negative and the maximum value of $y$, when $x = h$ is $k$ i.e. the parabola falls on either side and has highest point as $\left(h , k\right)$. For example, $y = - 3 {x}^{2} - 6 x + 9 = - 3 {\left(x + 1\right)}^{2} + 12$ and graph of $y = - 3 {x}^{2} - 6 x + 9$ has maxima at $\left(- 1 , 12\right)$ and observe that $\left(- 1 , 12\right)$ appears as a vertex.
graph{-3x^2-6x+9 [-5.433, 2.567, -6, 14]}