Y-INTERCEPT
[tex]y = 5x^2 - 16x + 10[/tex]
The y-intercept is where the equation/curve/parabola cosses the y-axis.
The y-axis is where x = 0. (The x-axis is where y = 0)
To find the y-intercept:
[tex]\text{y-axis} \rightarrow \text{x = 0} \rightarrow y = 5(0)^2 -16(0) + 10 = 10[/tex]
The y-intercept must be at (0, 10)
X-INTERCEPT (ROOTS/SOLUTIONS)
[tex]y = 5x^2 - 16x + 10\\\text{make it equal 0}\\y = 0\\\therefore 5x^2 - 16x + 10 = 0[/tex]
We need to use the quadratic formula
The quadratic formula helps us find what values of [tex]x[/tex] make the equation = 0
Quadratic formula: [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-(-16) + \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\\x = \frac{16 + \sqrt{256-200}}{10}\\x = \frac{16 + \sqrt{56}}{10}\\x = \frac{16 + 2\sqrt{14}}{10}\\x = \frac{8 + \sqrt{14}}{5}\\\\\\x=\frac{-(-16) - \sqrt{(-16)^2-4(5)(10)}}{2(5)}\\\text{doing the same thing...}\\\text{end up with...}\\x = \frac{8 - \sqrt{14}}{5}\\[/tex]
The x-intercepts are at:
[tex](\frac{8 + \sqrt{14}}{5}, 0)\\(\frac{8 - \sqrt{14}}{5}, 0)[/tex]
