What is the y-intercept for the function?

Mathematics

2 answers:

Serjik [45]3 years ago

7 0

Answer:

The y -intercept of a graph is the point where the graph crosses the y -axis. (Because a function must pass the vertical line test , a function can have at most one y -intercept . ) The y -intercept is often referred to with just the y -value.

Step-by-step explanation:

Y - intercept represents the position of a point on y-axis or when a line passes through y-axis, it, actually, passes through a point on y-axis. And that point is called the y intercept. It is usually represented by and as a point, it is represented as.

It's called the "y intercept" and it's the y value of the point where the line intersects the y- axis. For this line, the y-intercept is "negative 1." You can find the y-intercept by looking at the graph and seeing which point crosses the y axis. This point will always have an x coordinate of zero.

Hoped this helped! :D

bagirrra123 [75]3 years ago

7 0

Y intercept is 8.5 for this function

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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:

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