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

Answer:

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

Step-by-step explanation:

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

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

Using the quadratic equation:

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

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

$x=\frac{-\left(-8\right)\pm\sqrt{\left(-8\right)^2-4\cdot \:4\left(-3\right)}}{2\cdot \:4}$

$x=\frac{8\pm\sqrt{112}}{2\cdot \:4}$

But note that:

$\sqrt{117} =\sqrt{2^4\cdot \:7}=4\sqrt{7}$

So,

$x=\frac{8\pm4\sqrt{7}} {8}$

$x=\frac{2\pm\sqrt{7}} {2}$

Zeros/Roots:

$x_{1}=\frac{2+\sqrt{7}} {2}$

$x_{2}=\frac{2-\sqrt{7}} {2}$

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

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