Question

IM STUCK ! how do I determine the zeros if my x intercepts from the quadratic formula is this ! (in the picture) what do I do to find the y intercept ?!

Answer 1

The y-intercept is -3, but consider that 'a' is 4 and not 1, in your solving. Hope, I helped you.

Explanation:

First, let's solve the quadratic equation given, and find the x-intercepts:

`4x^2-8x-3=0`

Using the quadratic equation:

`$x=(-b\pm √(b^2-4ac))/(2a)$`

`a=4, b=-8, c=-3`

`$x=(-\left(-8\right)\pm√(\left(-8\right)^2-4\cdot \:4\left(-3\right)))/(2\cdot \:4)$`

`$x=(8\pm√(112))/(2\cdot \:4)$`

But note that:

`√(117) =√(2^4\cdot \:7)=4√(7)`

So,

`$x=\frac{8\pm4√(7)} {8}$`

`$x=\frac{2\pm√(7)} {2}$`

Zeros/Roots:

`$x_(1)=\frac{2+√(7)} {2}$`

`$x_(2)=\frac{2-√(7)} {2}$`

The y-intercept is when x is equal to 0, so:

`y=4x^2-8x-3\\y=4(0)^2-8(0)-3\\y=4 \cdot 0 - 8 \cdot 0 - 3\\y=-3`