Answer:
A system of equations can intersect at no points; this is when the lines are parallel, which means they have the same slope and different y-intercept. A system of equations can intersect at one point; this is when the lines have different slopes.
Step-by-step explanation:
This gives you three simultaneous equations:
6 = a + c
7 = 4a + c
1 = c
<u>c = 1
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If c =1,
6 = a + 1
<u>a = 5
</u><u /><u />
This doesn't work in the second equation, so the quadratic that goes through these points is not in the form y = ax^2 + bx + c
Was there supposed to be a b in the equation?
It's a slope-intercept form where a slope = -1.5 and y-intercept = 3.
x - intercept: y = 0
Therefore we have the equation:
-1.5x + 3 = 0 |-3
-1.5x = -3 |:(-1.5)
x = 2
Answer: x-intercept = 2, y-intercept = 3
X²(x - 4) +4 (x - 4)
(x² + 4) (x - 4)
First find the common terms that can enter into both x³ and 4x² then write its down in this case it’s x² that can enter x³ leaving only x _since x³/x² = subtract of the indices. x² will also enter 4x² leaving only four hence you having x² (x - 4)
then do the same for the next pair of terms giving you 4 that can enter into both 4 and 16
Leaving you with +4 (x - 4)
Now you can put the common terms together like so (x² + 4) and choose get one of the other two which are the same= (x - 4)
= (x² + 4) (x - 4)
