# How do you identify the important parts of y= 2x^2+7x-21 to graph it?

Oct 11, 2015

y intercept = -21 ; shape is an upward horse shoe with the minimum at $\left(1 \frac{3}{4} , - 1 \frac{3}{4}\right)$ Crosses the x axis at y=0 so factorise and equate to 0

#### Explanation:

equation standard form is $y = a {x}^{2} + b x + c$

in this case:

$+ {x}^{2}$ is upwards horse shoe shape
$- {x}^{2}$ is downward horse shoe shape
So this an upwards with a minimum y value but goes on increasing from there to infinity.

y intercept at $x = 0$ so by substitution $y = - 21$
which is the value of the constant c.

The value of $x$ at the minimum may be found by completing the square ( changing the way the equation looks without changing its intrinsic value). We need to end up with both the x^2 and the x part together inside the brackets. In this case we have:

write $y = 2 {x}^{2} + 7 x - 21$ as:

$y = 2 {\left({x}^{2} + \frac{7}{2}\right)}^{2} - 21 - \left(\text{correction bit of} {\left(\frac{7}{2}\right)}^{2}\right)$

the correction is to compensate for the constant we have introduced. We still have the -21 but we have introduced ${\left(\frac{7}{2}\right)}^{2}$ due to $\frac{7}{2} \times \frac{7}{2}$ from $\left({x}^{2} + \frac{7}{2}\right) \left({x}^{2} + \frac{7}{2}\right)$

Now we look inside the brackets at the $\frac{7}{2} x$ part
The minimum occurs at ${x}_{\min} = - \frac{1}{2} \times \frac{7}{2} = - 1 \frac{3}{4}$

To find ${y}_{\min}$ we substitute the found value for ${x}_{\min}$ in the original equation.

Oct 11, 2015

ABR again could not correct my answer. The value of ${x}_{\min}$ is 1/ 3/4# Sorry about that!!!