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Complete the T chart of values for the Quadratic: $\newline$ When you have checked your quiz and know that your T-Chart is correct, graph this Quadratic and upload it (along with the others) on the Graph Upload Assignment for points $\newline$ $y=x^{2}+5$ $\newline$ $\newline$ $x$ $\newline$ $y$ $\newline$ $\newline$ $0$ $\newline$ $\newline$ $1$ $\newline$ $\newline$ $-1$ $\newline$ $\newline$ $2$ $\newline$ $-2$ $\newline$ $3$ $\newline$ $-3$

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Find the magnitude of a vector

Full solution

Q. Complete the T chart of values for the Quadratic:

\newline

When you have checked your quiz and know that your T-Chart is correct, graph this Quadratic and upload it (along with the others) on the Graph Upload Assignment for points

\newline

y=x^{2}+5

\newline

\newline

x

\newline

y

\newline

\newline

0

\newline

\newline

1

\newline

\newline

-1

\newline

\newline

2

\newline

-2

\newline

3

\newline

-3

Substitute $x$ with $0$ : To find the value of $y$ when $x = 0$ , we substitute $x$ with $0$ in the equation $y = x^2 + 5$ . $\newline$ Calculation: $y = 0^2 + 5 = 0 + 5 = 5$
Substitute $x$ with $1$ : To find the value of $y$ when $x = 1$ , we substitute $x$ with $1$ in the equation $y = x^2 + 5$ . $\newline$ Calculation: $y = 1^2 + 5 = 1 + 5 = 6$
Substitute $x$ with $-1$ : To find the value of $y$ when $x = -1$ , we substitute $x$ with $-1$ in the equation $y = x^2 + 5$ . $\newline$ Calculation: $y = (-1)^2 + 5 = 1 + 5 = 6$
Substitute $x$ with $2$ : To find the value of $y$ when $x = 2$ , we substitute $x$ with $2$ in the equation $y = x^2 + 5$ . $\newline$ Calculation: $y = 2^2 + 5 = 4 + 5 = 9$
Substitute $x$ with $-2$ : To find the value of $y$ when $x = -2$ , we substitute $x$ with $-2$ in the equation $y = x^2 + 5$ . $\newline$ Calculation: $y = (-2)^2 + 5 = 4 + 5 = 9$
Substitute $x$ with $3$ : To find the value of $y$ when $x = 3$ , we substitute $x$ with $3$ in the equation $y = x^2 + 5$ . $\newline$ Calculation: $y = 3^2 + 5 = 9 + 5 = 14$
Substitute $x$ with $-3$ : To find the value of $y$ when $x = -3$ , we substitute $x$ with $-3$ in the equation $y = x^2 + 5$ . $\newline$ Calculation: $y = (-3)^2 + 5 = 9 + 5 = 14$

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