MATH HELP PLEASE 50 POINTS:
1 answer:
Answer:
where a>0.
To graph the the polynomial, begin in the left top of quadrant 2. Then draw downwards to the first real zero on the x-axis at -2. Cross the x-axis and then curve back up to 1/2 on the x-axis. Cross through again and curve back down to cross for the last time at 3 on the x-axis. The graph then ends going down towards the right in quadrant 4. It forms an s shape.
Step-by-step explanation:
The real zeros are the result of setting each factor of the polynomial to zero. By reversing this process, we find:
- zero -2 is factor (x+2)
- zero 1/2 is factor (2x-1)
- zero 3 is factor (x-3)
We write them together with an unknown leading coefficient a which is negative so -a.
where a>0
The leading coefficient of a polynomial determines the direction of the graph's end behavior.
- A positive leading coefficient has the end behavior point up when an even degree and point opposite directions when an odd degree with the left down and the right up.
- A negative leading coefficient has the end behavior point down when an even degree and point opposite directions when an odd degree with the left up and the right down.
- This graph has all odd multiplicity. The graph will cross through the x-axis each time at its real zeros.
To graph the the polynomial, begin in the left top of quadrant 2. Then draw downwards to the first real zero on the x-axis at -2. Cross the x-axis and then curve back up to 1/2 on the x-axis. Cross through again and curve back down to cross for the last time at 3 on the x-axis. The graph then ends going down towards the right in quadrant 4. It forms an s shape.
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Answer:
Step-by-step explanation:
A line perpendicular to the given line has a slope that is the negative inverse of the reference line.
Rewrite the given equation in the format of y=mx+b, where mi is the slope and b is the y-intercept (the value of y when x = 0.
2x + 3y = 4
3y=-2x+4
y = -(2/3)X + (4/3)
The reference slope is -(2/3). The negative inverse is (3/2), which will be the slope of a perpendicular line. We can write the new line as:
y = (3/2)x + b
Any value of b will still result in a line that is perpendicular. But we want a value of b that will shift the line so that it intersects the point (-3,-5). Simply enter this point in the above equation and solve for b.
y = (3/2)x + b
-5 = (3/2)(-3) + b
-5 = -(9/2) + b
-5 = -4.5 + b
b = - 0.5
The equation of the line that is perpendicular to 2x + 3y = 4 and includes point (-3,-5) is
y = (3/2)x - 0.5
