# Where does the graph of y = 2x^2 + x − 15 cross the x-axis?

Jun 2, 2015

Cutting the $x$ axis means $y = 0$
Which means 2x²+x-15=0

We are going to seek the $\Delta$ :
The equation is of the form ax²+bx+c=0
$a = 2$ ; $b = 1$ ; $c = - 15$

Delta=b²-4ac
Delta=1²-4*2*(-15)
$\Delta = 1 + 120$
$\Delta = 121$ ( $= \sqrt{11}$ )

${x}_{1} = \frac{- b - \sqrt{\Delta}}{2 a}$

${x}_{1} = \frac{- 1 - 11}{4}$

${x}_{1} = - \frac{12}{4}$

${x}_{1} = - 3$

${x}_{2} = \frac{- b + \sqrt{\Delta}}{2 a}$

${x}_{2} = \frac{- 1 + 11}{4}$

${x}_{2} = \frac{10}{4}$

${x}_{2} = \frac{5}{2}$

Thus, the function cuts the $x$ axis in $x = - 3$ and $x = \frac{5}{2}$

graph{2x^2+x-15 [-10, 10, -5, 5]}

Jun 2, 2015

$y = 2 {x}^{2} + x - 15 = \left(2 x - 5\right) \left(x + 3\right)$

$y = 0$ when $x = \frac{5}{2}$ or $x = - 3$

so the graph crosses the x-axis at $\left(- 3 , 0\right)$ and $\left(\frac{5}{2} , 0\right)$