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Write the standard form of the quadratic equation modeled by the points shown in the table below.


x y

-1 14

0 7

1 4

2 5

3 10

y=?

Respuesta :

Answer:y =ax² + bx +c

1) Point (0,7)

7 = a*0² +b*0 +c

c = 7

y=ax² + bx + 7

2) Point (1,4)

4=a*1² + b*1 + 7, ----> 4 = a +b + 7, ------> a+b= - 3

3) Point (2, 5)

5=a*2² + b*2 + 7, ----> 5=4a+2b +7,---> -2=4a+2b, ----> -1=2a + b

4)  a+b= - 3, ----> b= -3 - a (substitute in the second equation)

    2a+b= -1

2a - 3 - a = -1, ----> a - 3 = -1, a =2

5) a+b= - 3

    2 + b = -3

b = -5

y=2x² - 5x + 7

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Step-by-step explanation:

Answer:

[tex]\displaystyle y = 2x^2 - 5x + 7[/tex]

Step-by-step explanation:

You can determine this by logic. Here is what I mean:

First off, [tex]\displaystyle [3, 10]:[/tex]

[tex]\displaystyle 10 = A[3]^2 - B[3] + 7 → 10 = A9 - B3 + 7[/tex]

Now, knowing that [tex]\displaystyle A9 - B3[/tex]has to equal 3, we can find a SEPARATE common multiple for 3 and 9, and for 9 is 18, and for 3 is 15. Since these two numbers differ to 3, we get this genuine statement:

[tex]\displaystyle 10 = 2[3]^2 - 5[3] + 7 → 10 = 2[9] - 15 + 7 → 10 = 18 - 15 + 7 → 10 = 10; -5 = B, 2 = A[/tex]

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[tex]\displaystyle [2, 5][/tex]

[tex]\displaystyle 5 = A[2]^2 - B[2] + 7 → 5 = A4 - B2 + 7[/tex]

Now, knowing that [tex]\displaystyle A4 - B2[/tex]has to equal −2, we can find a SEPARATE common multiple for 2 and 4, and for 4 is 8, and for 2 is 10. Since these two numbers differ to −2, we get this genuine statement:

[tex]\displaystyle 5 = 2[2]^2 - 5[2] + 7 → 5 = 2[4] - 10 + 7 → 5 = 8 - 10 + 7 → 5 = 5; -5 = B, 2 = A[/tex]

So far so good!

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[tex]\displaystyle [1, 4][/tex]

[tex]\displaystyle 4 = A[1]^2 - B[1] + 7 → 4 = A - B + 7[/tex]

Now, knowing that [tex]\displaystyle A - B[/tex]has to equal −3, we can find a SEPARATE common multiple for both ones, and for 1 is 2, and for 1 is 5. Since these two numbers differ to −3, we get this genuine statement:

[tex]\displaystyle 4 = 2[1]^2 - 5[1] + 7 → 4 = 2[1] - 5 + 7 → 4 = 2 - 5 + 7 → 4 = 4; -5 = B, 2 = A[/tex]

One more to go!

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[tex]\displaystyle [-1, 14][/tex]

[tex]\displaystyle 14 = A[-1]^2 - B[-1] + 7 → 14 = A + B + 7[/tex]

Now, knowing that [tex]\displaystyle A + B[/tex]has to equal 7, we can find a SEPARATE common multiple for both ones, and for 1 is 2, and for 1 is 5. Since these two numbers add to 7, we get this genuine statement:

[tex]\displaystyle 14 = 2[-1]^2 - 5[-1] + 7 → 14 = 2[1] + 5 + 7 → 14 = 2 + 5 + 7 → 14 = 14; -5 = B, 2 = A[/tex]

* As you know, we never had to use the y-intercept of [tex]\displaystyle [0, 7][/tex]because that is AUTOMATICALLY our C-value when x is set equaled to 0.

I am joyous to assist you anytime.