All Algebra II Resources
Example Questions
Example Question #58 : Transformations Of Linear Functions
Translate the equation down six units. What is the new equation?
Use distribution to simplify the equation.
The equation in slope-intercept form is:
Shift the equation down six units means subtract the y-intercept by six.
The answer is:
Example Question #59 : Transformations Of Linear Functions
For the function we will define a linear transformation such that . Find the slope and y-intercept of the inverse function of .
Use the given function to find the linear transformation defined by .
First understand that in order to write we have to take our function and evaluate it for and then add to the result.
Adding ,
Now that we have , compute the inverse . Conventionally we replace the notation with the notation and solve for
Therefore,
At this point it's conventional to interchange and to write the inverse since we want to express it as a function of
Therefore the slope is and the y-intercept is
Example Question #59 : Transformations Of Linear Functions
Transform the equation into slope-intercept form.
In order to take an action from standard form to slope-intercept form you want to make it of the form:
where is the y-intercept (constant/number without a variable attached to it)
is the slope and coefficient of the term
and the equation is set equal to making it easier to plot graphically.
Given:
I. Isolate on one side of the equation. This is done by shifting either over to the other side of equation or the term and the constant to the other side of the equation. It is generally preferable to shift it to make so the term is positive by itself to simply operations and sign mixups. So in this case both and would be subtracted from both sides of the equation leaving:
II. Now that is isolated by itself you want to simplify the equation so there is no coefficient other than attached to , so in this case it'd mean dividing both sides by leaving:
III. Simplify the expression. If you are given fractions that are divisible by each other they can be simplified.
In the case is simplified to and is simplified to leaving the final answer of:
Example Question #1 : Quadratic Functions
Write a quadratic equation having as the vertex (vertex form of a quadratic equation).
The vertex form of a quadratic equation is given by
Where the vertex is located at
giving us .
Example Question #1 : Quadratic Functions
What are the -intercepts of the equation?
There are no -intercepts.
To find the x-intercepts of the equation, we set the numerator equal to zero.
Example Question #1 : Quadratic Functions
What is the minimum possible value of the expression below?
The expression has no minimum value.
We can determine the lowest possible value of the expression by finding the -coordinate of the vertex of the parabola graphed from the equation . This is done by rewriting the equation in vertex form.
The vertex of the parabola is the point .
The parabola is concave upward (its quadratic coefficient is positive), so represents the minimum value of . This is our answer.
Example Question #924 : Algebra Ii
Find the coordinates of the vertex of this quadratic function:
Vertex of quadratic equation is given by .
For ,
,
so the coordinate of vertex is .
Example Question #925 : Algebra Ii
What are the x-intercepts of the graph of ?
Assume y=0,
,
Example Question #2 : Quadratic Functions
Find the vertex of the parabola given by the following equation:
In order to find the vertex of a parabola, our first step is to find the x-coordinate of its center. If the equation of a parabola has the following form:
Then the x-coordinate of its center is given by the following formula:
For the parabola described in the problem, a=-2 and b=-12, so our center is at:
Now that we know the x-coordinate of the parabola's center, we can simply plug this value into the function to find the y-coordinate of the vertex:
So the vertex of the parabola given in the problem is at the point
Example Question #1 : Parabolic Functions
Give the minimum value of the function .
This function does not have a minimum.
This is a quadratic function. The -coordinate of the vertex of the parabola can be determined using the formula , setting :
Now evaluate the function at :