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3 years ago

Find the x-intercept for y = 3x^2 + 6x − 10

Mathematics

1 answer:

azamat3 years ago

5 0

Answer:

x-intercepts: (-3.08, 0) and (1.08, 0)

Step-by-step explanation:

Given:

The function is given as:

$y=3x^2+6x-10$

In order to find the x-intercept, we need to equate the given function to 0 as x-intercept is the point where the 'y' value is 0. So,

$y=0\\3x^2+6x-10=0$

Now, this is a quadratic equation of the form $ax^2+bx+c=0$

We find the solution using the quadratic formula,

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

Here, $a=3,b=6,c=-10$

Now, the solutions are:

$x=\frac{-6\pm \sqrt{6^2-4(3)(-10)}}{2(3)}\\\\x=\frac{-6\pm \sqrt{36+120}}{6}\\\\x=\frac{-6\pm \sqrt{156}}{6}\\\\x=\frac{-6}{6}-\frac{2\sqrt{39}}{6}\ or\ x=\frac{-6}{6}+\frac{2\sqrt{39}}{6}\\\\x=-1-2.08\ or\ x=-1+2.08\\\\x=-3.08\ or\ x=1.08$

Therefore, the x-intercepts are (-3.08, 0) and (1.08, 0)

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Normal vector tangent to plane is given by:

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$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

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