(2x−5)2−.5x<(2x−1)(2x+1)−15
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Y =ax² + bx +c
1) Point (0,7)
7 = a*0² +b*0 +c
c = 7
y=ax² + bx + 7
2) Point (1,4)
4=a*1² + b*1 + 7, ----> 4 = a +b + 7, ------> a+b= - 3
3) Point (2, 5)
5=a*2² + b*2 + 7, ----> 5=4a+2b +7,---> -2=4a+2b, ----> -1=2a + b
4) a+b= - 3, ----> b= -3 - a (substitute in the second equation)
2a+b= -1
2a - 3 - a = -1, ----> a - 3 = -1, a =2
5) a+b= - 3
2 + b = -3
b = -5
y=2x² - 5x + 7
1) Point (0,7)
7 = a*0² +b*0 +c
c = 7
y=ax² + bx + 7
2) Point (1,4)
4=a*1² + b*1 + 7, ----> 4 = a +b + 7, ------> a+b= - 3
3) Point (2, 5)
5=a*2² + b*2 + 7, ----> 5=4a+2b +7,---> -2=4a+2b, ----> -1=2a + b
4) a+b= - 3, ----> b= -3 - a (substitute in the second equation)
2a+b= -1
2a - 3 - a = -1, ----> a - 3 = -1, a =2
5) a+b= - 3
2 + b = -3
b = -5
y=2x² - 5x + 7