Solved on Dec 14, 2023

Find the value of $\mathrm{c}$ in the standard form of a quadratic function $y = ax^2 + bx + c$ given the table of values.

STEP 1

Assumptions
1. We are given a table of values for a function $y = f(x)$ .
2. We need to determine if these values fit a quadratic function of the form $y = ax^2 + bx + c$ .
3. If it is a quadratic function, we need to find the value of $c$ .
4. We are given multiple-choice options for the value of $c$ .

STEP 2

To determine if the function is quadratic and to find the value of $c$ , we will use the given values of $x$ and $y$ to create a system of equations. We will use three points to create three equations.

STEP 3

Let's use the points $(1, 6)$ , $(2, 28)$ , and $(3, 58)$ to create our system of equations. We will substitute these points into the standard form of a quadratic function $y = ax^2 + bx + c$ .

STEP 4

Substitute the first point $(1, 6)$ into the quadratic equation:
$6 = a(1)^2 + b(1) + c$

STEP 5

Simplify the equation from STEP_4:
$6 = a + b + c$

STEP 6

Substitute the second point $(2, 28)$ into the quadratic equation:
$28 = a(2)^2 + b(2) + c$

STEP 7

Simplify the equation from STEP_6:
$28 = 4a + 2b + c$

STEP 8

Substitute the third point $(3, 58)$ into the quadratic equation:
$58 = a(3)^2 + b(3) + c$

STEP 9

Simplify the equation from STEP_8:
$58 = 9a + 3b + c$

STEP 10

We now have a system of three equations:
\begin{align} a + b + c &= 6 \quad \text{(Equation 1)} \\ 4a + 2b + c &= 28 \quad \text{(Equation 2)} \\ 9a + 3b + c &= 58 \quad \text{(Equation 3)} \end{align}

STEP 11

Subtract Equation 1 from Equation 2 to eliminate $c$ and find an equation in terms of $a$ and $b$ .

STEP 12

Perform the subtraction:
\begin{align} (4a + 2b + c) - (a + b + c) &= 28 - 6 \\ 3a + b &= 22 \quad \text{(Equation 4)} \end{align}

STEP 13

Subtract Equation 1 from Equation 3 to eliminate $c$ and find another equation in terms of $a$ and $b$ .

STEP 14

Perform the subtraction:
\begin{align} (9a + 3b + c) - (a + b + c) &= 58 - 6 \\ 8a + 2b &= 52 \quad \text{(Equation 5)} \end{align}

STEP 15

Divide Equation 5 by 2 to simplify it:
$4a + b = 26 \quad \text{(Equation 6)}$

STEP 16

Subtract Equation 4 from Equation 6 to solve for $a$ .

STEP 17

Perform the subtraction:
\begin{align} (4a + b) - (3a + b) &= 26 - 22 \\ a &= 4 \end{align}

STEP 18

Substitute $a = 4$ into Equation 4 to solve for $b$ .

STEP 19

Perform the substitution:
$3(4) + b = 22$

STEP 20

Solve for $b$ :
$12 + b = 22$
$b = 22 - 12$
$b = 10$

STEP 21

Now that we have $a$ and $b$ , we can substitute these values into Equation 1 to solve for $c$ .

STEP 22

Perform the substitution:
$4 + 10 + c = 6$

STEP 23

Solve for $c$ :
$14 + c = 6$
$c = 6 - 14$
$c = -8$

STEP 24

The value of $c$ is $-8$ , which means the correct answer is C. -8.
The function $y = f(x)$ is indeed a quadratic function, and the number used for " $\mathrm{c}$ " in the standard form of the quadratic function is $-8$ .

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Find the value of c\mathrm{c}c in the standard form of a quadratic function y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c given the table of values.

STEP 1

STEP 2

To determine if the function is quadratic and to find the value of ccc, we will use the given values of xxx and yyy to create a system of equations. We will use three points to create three equations.

STEP 3

Let's use the points (1,6)(1, 6)(1,6), (2,28)(2, 28)(2,28), and (3,58)(3, 58)(3,58) to create our system of equations. We will substitute these points into the standard form of a quadratic function y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c.

STEP 4

Substitute the first point (1,6)(1, 6)(1,6) into the quadratic equation:6=a(1)2+b(1)+c6 = a(1)^2 + b(1) + c6=a(1)2+b(1)+c

STEP 5

Simplify the equation from STEP_4:6=a+b+c6 = a + b + c6=a+b+c

STEP 6

Substitute the second point (2,28)(2, 28)(2,28) into the quadratic equation:28=a(2)2+b(2)+c28 = a(2)^2 + b(2) + c28=a(2)2+b(2)+c

STEP 7

Simplify the equation from STEP_6:28=4a+2b+c28 = 4a + 2b + c28=4a+2b+c

STEP 8

Substitute the third point (3,58)(3, 58)(3,58) into the quadratic equation:58=a(3)2+b(3)+c58 = a(3)^2 + b(3) + c58=a(3)2+b(3)+c

STEP 9

Simplify the equation from STEP_8:58=9a+3b+c58 = 9a + 3b + c58=9a+3b+c

STEP 10

We now have a system of three equations:\begin{align*} a + b + c &= 6 \quad \text{(Equation 1)} \\ 4a + 2b + c &= 28 \quad \text{(Equation 2)} \\ 9a + 3b + c &= 58 \quad \text{(Equation 3)} \end{align*}

STEP 11

Subtract Equation 1 from Equation 2 to eliminate ccc and find an equation in terms of aaa and bbb.

STEP 12

Perform the subtraction:\begin{align*} (4a + 2b + c) - (a + b + c) &= 28 - 6 \\ 3a + b &= 22 \quad \text{(Equation 4)} \end{align*}

STEP 13

Subtract Equation 1 from Equation 3 to eliminate ccc and find another equation in terms of aaa and bbb.

STEP 14

Perform the subtraction:\begin{align*} (9a + 3b + c) - (a + b + c) &= 58 - 6 \\ 8a + 2b &= 52 \quad \text{(Equation 5)} \end{align*}

STEP 15

Divide Equation 5 by 2 to simplify it:4a+b=26(Equation 6)4a + b = 26 \quad \text{(Equation 6)}4a+b=26(Equation 6)

STEP 16

Subtract Equation 4 from Equation 6 to solve for aaa.

STEP 17

Perform the subtraction:\begin{align*} (4a + b) - (3a + b) &= 26 - 22 \\ a &= 4 \end{align*}

STEP 18

Substitute a=4a = 4a=4 into Equation 4 to solve for bbb.

STEP 19

Perform the substitution:3(4)+b=223(4) + b = 223(4)+b=22

STEP 20

Solve for bbb:12+b=2212 + b = 2212+b=22b=22−12b = 22 - 12b=22−12b=10b = 10b=10

STEP 21

Now that we have aaa and bbb, we can substitute these values into Equation 1 to solve for ccc.

STEP 22

Perform the substitution:4+10+c=64 + 10 + c = 64+10+c=6

STEP 23

Solve for ccc:14+c=614 + c = 614+c=6c=6−14c = 6 - 14c=6−14c=−8c = -8c=−8

STEP 24

The value of ccc is −8-8−8, which means the correct answer is C. -8.The function y=f(x)y = f(x)y=f(x) is indeed a quadratic function, and the number used for "c\mathrm{c}c" in the standard form of the quadratic function is −8-8−8.

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Find the value of $\mathrm{c}$ in the standard form of a quadratic function $y = ax^2 + bx + c$ given the table of values.

To determine if the function is quadratic and to find the value of $c$ , we will use the given values of $x$ and $y$ to create a system of equations. We will use three points to create three equations.

Let's use the points $(1, 6)$ , $(2, 28)$ , and $(3, 58)$ to create our system of equations. We will substitute these points into the standard form of a quadratic function $y = ax^2 + bx + c$ .

Substitute the first point $(1, 6)$ into the quadratic equation:
$6 = a(1)^2 + b(1) + c$

Simplify the equation from STEP_4:
$6 = a + b + c$

Substitute the second point $(2, 28)$ into the quadratic equation:
$28 = a(2)^2 + b(2) + c$

Simplify the equation from STEP_6:
$28 = 4a + 2b + c$

Substitute the third point $(3, 58)$ into the quadratic equation:
$58 = a(3)^2 + b(3) + c$

Simplify the equation from STEP_8:
$58 = 9a + 3b + c$

We now have a system of three equations:
\begin{align} a + b + c &= 6 \quad \text{(Equation 1)} \\ 4a + 2b + c &= 28 \quad \text{(Equation 2)} \\ 9a + 3b + c &= 58 \quad \text{(Equation 3)} \end{align}

Subtract Equation 1 from Equation 2 to eliminate $c$ and find an equation in terms of $a$ and $b$ .

Perform the subtraction:
\begin{align} (4a + 2b + c) - (a + b + c) &= 28 - 6 \\ 3a + b &= 22 \quad \text{(Equation 4)} \end{align}

Subtract Equation 1 from Equation 3 to eliminate $c$ and find another equation in terms of $a$ and $b$ .

Perform the subtraction:
\begin{align} (9a + 3b + c) - (a + b + c) &= 58 - 6 \\ 8a + 2b &= 52 \quad \text{(Equation 5)} \end{align}

Divide Equation 5 by 2 to simplify it:
$4a + b = 26 \quad \text{(Equation 6)}$

Subtract Equation 4 from Equation 6 to solve for $a$ .

Perform the subtraction:
\begin{align} (4a + b) - (3a + b) &= 26 - 22 \\ a &= 4 \end{align}

Substitute $a = 4$ into Equation 4 to solve for $b$ .

Perform the substitution:
$3(4) + b = 22$

Solve for $b$ :
$12 + b = 22$
$b = 22 - 12$
$b = 10$

Now that we have $a$ and $b$ , we can substitute these values into Equation 1 to solve for $c$ .

Perform the substitution:
$4 + 10 + c = 6$

Solve for $c$ :
$14 + c = 6$
$c = 6 - 14$
$c = -8$

The value of $c$ is $-8$ , which means the correct answer is C. -8.
The function $y = f(x)$ is indeed a quadratic function, and the number used for " $\mathrm{c}$ " in the standard form of the quadratic function is $-8$ .