Can someone help me with this exercise?
1 answer:
Answer:
View image
Step-by-step explanation:
You treat inequality questions just like normal equation with an equal sign. You only have to worry about the inequality when you do the shading part.
Your first goal is to solve for y, which I did at the top.
And remember that when you divide/multiply by a negative number you have to flip the inequality sign.
So the first equation is already in term of y. (y < 2x + 4)
So you only have to solve for y for -3x - 2y ≥ 6.
I solved that and got y ≤ -3/2x - 3.
Now you have 2 equations. y < 2x + 4 and y ≤ -3/2x - 3
First you gotta treat it like a normal equation and graph them.
So i graphed y = 2x + 4 and y = -3/2x - 3. I assume you already know how do do that.
btw. < and > have dotted lines, while ≤ and ≥ have solid lines.
Now for the shading, you gotta look at the inequality sign.
y < 2x + 4 has a less than sign, so you have to shade everything underneath the line you graphed.
y ≤ -3/2x - 3 is also less than, so you have to shade everything underneath that line as well.
So in the future, you you get a > or ≥ then you have to shade above the graph.
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Answer:
"complete the square" to put in vertex form
Step-by-step explanation:
It may be helpful to consider the square of a binomial:
(x +a)² = x² +2ax +a²
The expression x² +x +1 is in the standard form of the expression on the right above. Comparing the coefficients of x, we see ...
2a = 1
a = 1/2
That means we can write ...
(x +1/2)² = x² +x +1/4
But we need x² +x +1, so we need to add 3/4 to the binomial square in order to make the expressions equal:
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Another way to consider this is ...
x² +bx +c
= x² +2(b/2)x +(b/2)² +c -(b/2)² . . . . . . rewrite bx, add and subtract (b/2)²*
= (x +b/2)² +(c -(b/2)²)
for b=1, c=1, this becomes ...
x² +x +1 = (x +1/2)² +(1 -(1/2)²)
= (x +1/2)² +3/4
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* This process, "rewrite bx, add and subtract (b/2)²," is called "completing the square"—especially when written as (x-h)² +k, a parabola with vertex (h, k).