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3 years ago

8

A Polynomial Identity is True when you work the problem out and both sides are equal to each other

1 answer:

USPshnik [31]3 years ago

4 0

answer

it is true

step by step explanation

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3 years ago

Find the quadratic function y=ax^2 + bx + c whose graph passes through the given points. (-3,37), (2,-8), (-1,13)

melamori03 [73]

The quadratic function is $y=1x^{2} + (-8)x + 4$

Step-by-step explanation:

The quadratic function given is $ax^{2} + bx + c=y$

and same quadratic function is passes through (-3,37), (2,-8), (-1,13)

Replacing points one by one

we get,

For (-3,37) :

$a(-3)^{2} + b(-3) + c=37$

$9a + -3b + c=37$ =<em> equation 1 </em>

For (2,-8) :

$a(2)^{2} + b(2) + c=(-8)$

$4a + 2b + c=(-8)$ = <em>equation 2</em>

For (-1,13)

$a(-1)^{2} + b(-1) + c=(13)$

$a + -1b + c=13$ = <em>equation 3</em>

Solving the linear equation to get values of a,b,c

Subtract equation 2 with equation 3

we get, $(4a + 2b + c)-(a + -1b + c)=(-8)-13$

$(3a + 3b )=(-21)$

$(a + b )=(-7)$ = <em>equation 4</em>

Now, Subtract equation 1 with equation 2

we get, $(9a + -3b + c)-(4a + 2b + c)=(37)-(-8)$

$(5a - 5b )=(45)$

$(a - b )=(9)$ = <em>equation 5</em>

Now, Add equation 4 with equation 5

we get, $(a + b)+(a - b)=(-7)+(9)$

$(2a - 0b )=(2)$

$(a)=1$

Replacing value of a in <em>equation 5</em>

$(a - b )=(9)$

$(1 - b )=(9)$

$(b)=(-8)$

Replacing value of a and b in <em>equation 1</em>

$9a + -3b + c=37$

$9(1) + -3(-8) + c=37$

$9 + 24 + c=37$

$c=4$

Thus,

The quadratic function $y=1x^{2} + (-8)x + 4$

3 0

4 years ago