Question

−4x^2 − 16x + 2 x-intercepts

Answer 1

x-intercepts: x values when the function f(x) = y = 0

you can use the quadratic equation on your polynomial of the form ax^2 + bx + c

a = -4; b = -16; c = 2

quadratic equation:

\[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]

Answer 2

let me know if i should continue or if this is enough info

Answer 3

can you help me get the intercepts. I tried everything and it was still wrong

Answer 4

\[x = \frac{ -(-16) \pm \sqrt{(-16)^2 - 4(-4)(2)} }{ 2(-4) } = \frac{ 16 \pm \sqrt{256 + 32} }{ -8 }\]\[x = \frac{ 16 \pm \sqrt{288} }{ -8 } = \frac{ 16 \pm 12\sqrt{2} }{ -8 } = -2 \pm \frac{ -3 }{ 2 } \sqrt{2}\]

so the intercepts are:

\[-2 - \frac{ 3 }{ 2 } \sqrt{2}\]and\[-2 + \frac{ 3 }{ 2 } \sqrt{2}\]

Answer 5

omg thank you so much

Answer 6

glad i could help :)