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shutvik [7]
3 years ago
15

For 1-3, use the graph of the quadratic function.

Mathematics
1 answer:
Nataly_w [17]3 years ago
4 0

Answer:

Step-by-step explanation:

From the picture attached,

Function given is a parabola, therefore, function is a quadratic function.

Domain of the function : x-values(input values) on the graph

                                      : -∞ ≤ x < ∞

Domain in set notation : {x | x ∈ All real numbers}

Domain in the form interval notation : (-∞, ∞)

End behavior of the function : As both the ends of the function are downwards towards the infinity,

                                                : As x → ∞, y → -∞ [Right end]

                                                : As x → -∞, y → -∞ [Left end]      

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[10] In the following given system, determine a matrix A and vector b so that the system can be represented as a matrix equation
irina1246 [14]

Answer:

y=-\frac{158}{579}

Step-by-step explanation:

To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:

A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]

And the vector B is formed with the solution of each equation of the system:b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]

To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called A_{2}.

A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]

The value of y using Cramer's rule is:

y=\frac{det(A_{2}) }{det(A)}

Find the value of the determinant of each matrix, and divide:

y==\frac{\left|\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right|}{\left|\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right|} =\frac{158}{-579}

y=-\frac{158}{579}

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4 years ago
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