What are the x-intercepts of the graph of the following equation ?

Mathematics

1 answer:

anzhelika [568]3 years ago

4 0

Answer:

<h3> x-intercepts: (-6, 0) , (2, 0)</h3>

Step-by-step explanation:

x-intercepts are points where y=0

$y = -x^2 - 4x + 12\\\\y=0\quad\iff\quad -x^2 - 4x + 12=0\\\\{}\qquad\quad\quad a=-1\,,\ b=-4\,,\ c=12\\\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(4)\pm\sqrt{(-4)^2-4\cdot(-1)\cdot12}}{2\cdot(-1)}=\dfrac{4\pm\sqrt{16+48}}{-2}\\\\x_1=\dfrac{4+\sqrt{64}}{-2}=\dfrac{12}{-2}=-6\ ,\quad x_2=\dfrac{4-8}{-2}=\dfrac{-4}{-2}=2$

x-intercepts: (-6, 0) , (2, 0)

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now we sub 
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4. x = first integer 
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7. 2x + 3y <= 6 -- put in y = mx + b form 
3y <= -2x + 6 
y <= -2/3x + 2 
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To find the x intercept, sub in 0 for y (change inequality sign to equal sign) 
2x + 3y = 6 
2x + 3(0) = 6 
2x = 6 
x = 3 
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Connect the points with a solid line because there is an equal sign in there. 
Shade below the line because it is less then. 

8. 3x + 6y = 8 
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y = 2x - 8 

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9. 7x + 3y - 2 + 6x + 1 + y^2 
y^2 + 3y + 13x - 1 

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slope(m) = (0 - 6) / (3 - 0) 
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14. I cannot see the problem 

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open circle on -2, closed circle on 1, shading in between 

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Step-by-step explanation:

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AveGali [126]

Answer with Step-by-step explanation:

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