Answer:
Find a polynomial function whose graph passes through (6,13), (9,-11), (0,5)
1 Answers
Assuming a quadratic, we have that
y = ax^2 + bx + c
Since (0,5) is on the graph, c =5
And we have the remaining system
a(9)^2 + b(9) + 5 = -11
a(6)^2 + b(6) + 5 = 13 simplify
81a + 9b = -16 multiply through by 6 ⇒ 486a + 54b = - 96 (1)
36a + 6b = 8 multiply through by -9 ⇒ -324a -54b = -72 (2)
Add (1) and (2)
162a = -168
a = -28/27
To find b we have
36 (-28/27) + 6b = 8
-112/3 + 6b = 8
⇒ b = 68/9
The function is
y = - (28/27)x^2 + (68/9)x + 5
The value of 4 becomes 10 times greater in the second number than the first.
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Answer:
a^x/y=1 x: 0
Step-by-step explanation: w.k.t, a^0=1( any variable raised to 0 is 1)
so, here the exponent is x/y which should have been 0 so that answer was 1.
