Andrew claims the initial value and y - intercept are the same thing on a graph. Is he correct?

We conclude that the initial value and y-intercept are the same thing on a graph.

Please check the attached graph of the equation y = 2x+1.

Step-by-step explanation:

We know that the initial value on a graph is basically the out-put value y of the point where the line meets or crosses the y-axis.

In other words, the initial value is the y-value or output of the point at x = 0

For example,

Let the equation

y = 2x+1

substitute x = 0

y = 2(0)+1

y = 0+1

y = 1

Thus, the initial value of the equation y = 2x+1 is: y = 1

Please check the attached graph of the equation y = 2x+1.

It is clear from the graph that at x = 0, the value of y = 1.

Thus, at y = 1, the line meets the y-axis.

Hence, the initial value of the line is: y = 1

Similarly, we know that the value of the y-intercept can be determined by setting x = 0 and determining the corresponding value of y.

For example,

Let the equation

y = 2x+1

substitute x = 0

y = 2(0)+1

y = 0+1

y = 1

Thus, the y-intercept of y = 2x+1 is y = 1.

Please check the attached graph of the equation y = 2x+1.

It is clear from the graph that at x = 0, the value of y = 1.

Therefore, the y-intercept of y = 2x+1 is y = 1.

Conclusion:

Therefore, we conclude that the initial value and y-intercept are the same thing on a graph.

Please check the attached graph of the equation y = 2x+1.

step-by-step explanation:

try remembering the steps

answer: 5100 just subtract

step-by-step explanation: