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

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Answer 1 · 2015-06-02T18:47:18

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

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

$Delta=b²-4ac$
$Delta=1²-4*2*(-15)$
$Delta=1+120$
$Delta=121$ ( $=sqrt11$ )

$x_1=(-b-sqrtDelta)/(2a)$

$x_1=(-1-11)/4$

$x_1=-12/4$

$x_1=-3$

$x_2=(-b+sqrtDelta)/(2a)$

$x_2=(-1+11)/4$

$x_2=10/4$

$x_2=5/2$

Thus, the function cuts the $x$ axis in $x=-3$ and $x=5/2$

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

Answer 2 · 2015-06-02T18:50:05

$y = 2x^2+x-15 = (2x-5)(x+3)$

$y=0$ when $x = 5/2$ or $x=-3$

so the graph crosses the x-axis at $(-3, 0)$ and $(5/2, 0)$