The graph of a quadratic function passes through the points (5,0), (-3,0) and (-5,80) We can model the graph by using a function in the form of y = ax^2 + bx + c, where "a", "b" and "c" represent real numbers.

Find the value for "a", "b", and "c".

a = 4, b = - 8, c = - 60

Step-by-step explanation:

If the graph passes through these points, these points must satisfy the equation of the locus of that function, for each x and y.

For the solution (5, 0),

f(x) = y = ax^2 + bx + c

for (x, y) = (5, 0)

=> 0 = a(5)^2 + b(5) + c

=> 0 = 25a + 5b + c ...(1)

For (x, y) = (-3, 0)

=> 0 = a(-3)^2 + b(-3) + c

=> 0 = 9a - 3b + c ...(2)

For (x, y) = (-5, 80)

=> 80 = a(-5)^2 + b(-5) + c

=> 80 = 25a - 5b + c ...(3)

Subtract (1) from (3) we get,-80 = 10b => -8 = b. Thus, eq(1) is 40 = 25a + c, eq(2) is - 24 = 9a + c

Subtract (1) from (2),

64 = 16a => 4 = a

Hence, in eq(1), 40 = 25(4) + c, - 60 = c

8

step-by-step explanation:

find the magnitude of the trig term  

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find the lower bound of the range.

tap for more

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find the upper bound of the range.

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y

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(

−

∞

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{

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step-by-step explanation: