Respuesta :
Answer:
x-intercepts: (-3.08, 0) and (1.08, 0)
Step-by-step explanation:
Given:
The function is given as:
[tex]y=3x^2+6x-10[/tex]
In order to find the x-intercept, we need to equate the given function to 0 as x-intercept is the point where the 'y' value is 0. So,
[tex]y=0\\3x^2+6x-10=0[/tex]
Now, this is a quadratic equation of the form [tex]ax^2+bx+c=0[/tex]
We find the solution using the quadratic formula,
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Here, [tex]a=3,b=6,c=-10[/tex]
Now, the solutions are:
[tex]x=\frac{-6\pm \sqrt{6^2-4(3)(-10)}}{2(3)}\\\\x=\frac{-6\pm \sqrt{36+120}}{6}\\\\x=\frac{-6\pm \sqrt{156}}{6}\\\\x=\frac{-6}{6}-\frac{2\sqrt{39}}{6}\ or\ x=\frac{-6}{6}+\frac{2\sqrt{39}}{6}\\\\x=-1-2.08\ or\ x=-1+2.08\\\\x=-3.08\ or\ x=1.08[/tex]
Therefore, the x-intercepts are (-3.08, 0) and (1.08, 0)
