Find the coordinates of the points of intersection of the parabola y=x^2+3x-4 and the line y=2x+2

1 answer:

6 0

You can set the y values equal to each other and solve the system:

x^2+3x-4=2x+2
x^2+x-6=0
(x+3)(x-2)=0
x=-3,2

So, these are the x values of the intersection. To get the y values, you can just plug them back into either formula:

y=2(-3)+2
y=-4

y2(2)+2
y=6

So the coordinates are (-3,-4) and 2,6).

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2/6 + 1/8 A) 11/24 B) 3/14 C) 5/24 D) 8/18

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So with these two fractions you're gonna have to find the LCD (lowest common denominator) before you start adding them. With 6 and 8, the LCD is 24.

So with the numerator and denominator of $\frac{2}{6}$ multiply them by 4. It'll look like this $\frac{2}{6} * \frac{4}{4} = \frac{8}{24}$

As for $\frac{1}{8}$ , do the same except multiply by 3. It'll look like this: $\frac{1}{8} * \frac{3}{3} = \frac{3}{24}$ .

Now add the new fractions together: $\frac{8}{24} + \frac{3}{24} = \frac{11}{24}$

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