91Ó°ÊÓ

The first step to solving the given problem is function evaluation.
We need to find the values of the function at the specific points given in the problem.
The function given is: \[ f(x) = x^2 - 3x + 7 \]

Evaluate the function at the first point, which is \(x = -1\): \[ f(-1) = (-1)^2 - 3(-1) + 7 \]
Now, perform the calculations step-by-step: \[ f(-1) = 1 + 3 + 7 = 11 \]

Then, evaluate the function at the second point, \(x = 2\): \[ f(2) = (2)^2 - 3(2) + 7 \] Perform the calculations: \[ f(2) = 4 - 6 + 7 = 5 \]

After evaluating, you have the points (-1, 11) and (2, 5). This step is crucial as it sets the stage for finding the slope of the line containing these points.

Once you have the points, the next step is calculating the slope between them.
The slope, often denoted as \(m\), measures how steep a line is.
Use the slope formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Substitute the points \((-1, 11)\) and \((2, 5)\) into the formula: \[ m = \frac{5 - 11}{2 - (-1)} = \frac{ -6 }{ 3 } = -2 \]

Remember that the numerator is the difference in the \(y\) values and the denominator is the difference in the \(x\) values.
The slope calculation step tells us about the direction and steepness of the line.
In this case, the negative slope \(-2\) indicates that the line is decreasing (it goes down as you move to the right).

The final step is to write the equation of the secant line using the point-slope form.
The point-slope form of a line's equation is: \[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1)\) is a point on the line, and \(m\) is the slope.

Using the point \((-1, 11)\) and the slope \(-2\), substitute in to get: \[ y - 11 = -2(x - (-1)) \]
Simplify the equation step-by-step: \[ y - 11 = -2(x + 1) \]
further: \[ y - 11 = -2x - 2 \] then: \[ y = -2x + 9 \] Still simplifying, we get the final equation of the secant line, \[ y = -2x + 9 \] which is the equation you needed.

This step puts everything together, giving you the explicit equation for the secant line. The Point-Slope Form method is powerful and makes it easy to find the equation of the tangent or secant lines once you know the slope and a point on the line.