Standard quadratic equation .. y = a x^2 + b x + c
<span>parabola 'a' not equal to zero </span>
<span>a<0 parabola opens downward </span>
<span>a>0 parabola opens upward </span>
<span>when |a| >>0 the parabola is narrower </span>
<span>when |a| is close to zero , the parabola is flatter </span>
<span>when the constant is varied it only effects the vertical position of the parabola , the shape remains the same</span>
Answer:
-a+-2b+4c
Step-by-step explanation:
Combine like terms
-4a+ 3a= -a
b+ -3b= -2b
3c+ c= 4c
30/5= 6
6 - 2 = 4
Your number is four
Answer:
-2 and 12
Step-by-step explanation:
To find the two numbers which are 7 units away from 5, add 7 to 5 and subtract 7 from 5
5+7 = 12
5-7 = -2
The one that has positive 2 as the y intercept
