Question

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

Answer 1

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​) :

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

$9a + -3b + c=37$ = equation 1

For (2​,-8​) :

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

$4a + 2b + c=(-8)$ =  equation 2

For ​(-1​,13)

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

$a + -1b + c=13$ = equation 3

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

Subtract equation 2 with equation 3

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

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

$(a + b )=(-7)$  = equation 4

Now, Subtract equation 1 with equation 2

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

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

$(a - b )=(9)$  = equation 5

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 equation 5

$(a - b )=(9)$

$(1 - b )=(9)$

$(b)=(-8)$

Replacing value of a and b in equation 1

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

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

$9 + 24 + c=37$

$c=4$

Thus,

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