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icang [17]
3 years ago
13

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 ?!

Mathematics
1 answer:
N76 [4]3 years ago
5 0

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

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