tentukan fungsi kuadrat yang grafiknya memotong sumbu -x pada titik koordinat (4,0) dan (-3,0) serta malalui titik koordinat (2,-10)
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tentukan fungsi kuadrat yang grafiknya memotong sumbu -x pada titik koordinat (4,0) dan (-3,0) serta malalui titik koordinat (2,-10)
y = x² - x - 12
Step-by-step explanation:
Determine the quadratic function whose graph intersects the x-axis at the coordinates (4,0) and (-3,0) and through the coordinates (2, -10)
Let say y = ax² + bx + c
=> 0 = 16a + 4b + c Eq1
& 0 = 9a - 3b + c Eq2
-10 = 4a + 2b + c Eq3
Eq 1 - Eq2
=> 7a + 7b = 0
=> a = - b
=> 0 = 16a - 4a + c
=> c = -12a
-10 = 4a - 2a - 12a
=> -10a = -10
=> a = 1
b = -1
c = -12
y = x² - x - 12
Another way
y =(x - 4)(x - (-3))
=> y = (x-4)(x + 3)
=> y = x² - 4x + 3x - 12
=> y = x² - x - 12
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