Answer:
The function intersect the x-axis two times
Step-by-step explanation:
we have
[tex]y=-2x^{2}+3x+5[/tex]
To find the x-intercepts equate the equation to zero
so
[tex]0=-2x^{2}+3x+5[/tex]
[tex]-2x^{2}+3x+5=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]-2x^{2}+3x+5=0[/tex]
so
[tex]a=-2\\b=3\\c=5[/tex]
substitute in the formula
[tex]x=\frac{-3(+/-)\sqrt{3^{2}-4(-2)(5)}} {2(-2)}[/tex]
[tex]x=\frac{-3(+/-)\sqrt{49}} {-4}[/tex]
[tex]x=\frac{-3(+/-)7} {-4}[/tex]
[tex]x1=\frac{-3(+)7} {-4}=-1[/tex]
[tex]x2=\frac{-3(-)7} {-4}=2.5[/tex]
so
The function has two x-intercepts
therefore
The function intersect the x-axis two times
