# Let fxx2 2×35 enter the x-intercepts of the quadratic function in the boxes

## Enter the x-intercepts of the quadratic function in the boxes.

Enter the x-intercepts of the quadratic function in the boxes. The x-intercepts are the points where the graph of the quadratic function intersects the x-axis. These points can be found by setting f(x)=0 and solving for x.

## Use the Quadratic Formula to solve for the x-intercepts.

The Quadratic Formula is a very important equation for solving quadratic equations. It is written in the form:

x = [-b +/- sqrt(b^2-4ac)] / 2a

where a, b and c are the coefficients of the equation. In our case, we have:

a = 2
b = 35
c = 0

Plugging these values into the equation, we get:

x = [-35 +/- sqrt(35^2-4(2)(0))] / 2(2)

x = [-35 +/- sqrt(1225-0)] / 4

x = [-35 +/- sqrt(1225)] / 4

x = [-35 +/- 35] / 4 # take the square root of 1225 to get 35. This is called “factoring” the equation. We are simply writing 1225 as the product of two factors, 35 and 35. This is always possible when working with perfect squares.
x = [0 +/- 35] / 4

# Now we have two possibilities for x, either 0 + 35 or 0 – 35. These are our x-intercepts!

x = (0 + 35)/4 # this gives us our first x-intercept at x=8.75

x = (0 – 35)/4 # this gives us our second x-intercept at x=-8.75

## Enter the y-intercept of the quadratic function in the box.

The y-intercept of a quadratic function is the point where the graph of the function crosses the y-axis. It is the value of y when x = 0. To find the y-intercept, substitute 0 for x in the equation and solve for y.

## Use the Quadratic Formula to solve for the y-intercept.

If you need to find the y-intercept of a quadratic function, you can do so by using the Quadratic Formula. This formula is derived from the fact that a quadratic function can be written in the form y = ax^2 + bx + c, where a, b, and c are constants. The Quadratic Formula can be used to solve for any of these constants given the other two.

To use the Quadratic Formula, you first need to identify the values of a, b, and c in your equation. Once you have these values, you can plug them into the formula and solve for the y-intercept.

For example, let’s say you have the equation y = 2x^2 + 5x + 3. In this equation, a = 2, b = 5, and c = 3. Plugging these values into the Quadratic Formula gives us:

y = -(5)/(2)(2) + 3

y = -1/4 + 3

y = 2.75

Therefore, the y-intercept of this quadratic equation is 2.75.

## Graph the quadratic function.

To graph the quadratic function, you will need to find the x-intercepts of the function. These are the points where the graph of the function intersects the x-axis. To find the x-intercepts, you will need to solve the equation for y=0. This will give you two equations, one for each x-intercept. You can then solve these equations using the quadratic formula.

## Find the axis of symmetry.

The axis of symmetry is the line that divides a parabola into two identical halves. This line goes through the vertex of the parabola. To find the equation of the axis of symmetry, we take the average of the x-intercepts.

## Find the vertex.

After you have graphed the quadratic function, you can find the vertex by finding the point where the graph changes from concave up to concave down, or vice versa. In other words, the vertex is the point of inflection of the graph.

## Find the maximum or minimum value of the quadratic function.

There are several ways to find the maximum or minimum value of the quadratic function. One way is to use the Quadratic Formula. Another way is to take the derivative of the function and set it equal to 0 to find the critical points. Once you have found the critical points, you can plug them back into the original function to find the maximum or minimum value.