HOA P. answered  10/16/20
Excellent, hardworking and passionate Math Tutor
k(x)= x2+6x+5
x= 1 , k(1) = (1)^2 + 6(1) +5 = 1+6+5 = 12
(1,12)
x =2 k(2) = (2)^2 +6(2) +5 = 4+12+5 = 21
(2, 21)
McKinley C.
asked  10/16/20k(x)= x2+6x+5
A: (_______,______)
B: (_____,_______)
HOA P. answered  10/16/20
Excellent, hardworking and passionate Math Tutor
k(x)= x2+6x+5
x= 1 , k(1) = (1)^2 + 6(1) +5 = 1+6+5 = 12
(1,12)
x =2 k(2) = (2)^2 +6(2) +5 = 4+12+5 = 21
(2, 21)
Hi McKinley C.
For
k(x)= x2+6x+5
Another point simply would be the y intercept
(0,5)
Any value of x you want to plug into your equation is another point
When x = 1, y = 12
(1,12)
A= (0,5)
B=(1,12)
I hope you find this useful
        
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Bakar B.
Additionally, it is important to know where the vertex and x-intercepts are so those points aren't chosen: The x-intercepts occur where k(x) = 0; factoring the function and finding the roots (x-intercepts): k(x)= x^2+6x+5; k(x) = (x + 5)(x + 1) = 0; x = -5, x = -1; so neither A nor B should use these x-values. The vertex occurs midway between the x-intercepts, in this equation, because parabolas are always symmetrical. So x = -3 is also off-limits.10/16/20