All Precalculus Resources
Example Questions
Example Question #1 : Graph A Quadratic Function Using Intercepts, Vertex, And Axis Of Symmetry
Where does the vertex of the quadratic lie?
To find the vertex of a quadratic equation, you'll look to put the quadratic in the form , where is then the vertex. To get from the original equation to vertex form, you'll have to complete the square by looking at the terms that include and to turn that into a perfect square. Here you should see that with as the first two terms, you could have a perfect square if you could use . So to complete the square, you can express the given quadratic as:
Note that the +1 and -10 terms net out to the -9 that was in the original equation, so in this case you have not altered the value at all, but have merely reallocated numbers to fit vertex form (also note that there was no coefficient to the term, making the term in vertex form equal to 1).
From here you can factor the quadratic on the left to perfectly math vertex form:
This means that and , making the vertex .
Example Question #5 : Graph A Quadratic Function Using Intercepts, Vertex, And Axis Of Symmetry
What are the vertex and the line of symmetry for the quadratic ?
Vertex: (1, 3)
Line of Symmetry: x = 1
Vertex: (1, -5)
Line of Symmetry: x = 1
Vertex: (5, -1)
Line of Symmetry: x = 5
Vertex: (-3, -1)
Line of Symmetry: x = -3
Vertex: (-1, 5)
Line of Symmetry: x = 5
Vertex: (1, -5)
Line of Symmetry: x = 1
Note that solving for the x-coordinate of the vertex of a parabola also tells you its line of symmetry, so your job here is to put the quadratic into Vertex Form in order to find the vertex, which will give you the line of symmetry. Vertex form is , where is then the vertex. To get to that form, you will want to complete the square by looking at the and terms and determining which perfect square equation they belong to. To do that, separate those two terms from the -2 term, and then factor out the coefficient of 3:
Then note that the way to turn into a perfect square would be to add 1 to it to get to . Of course, you cannot just add one within the parentheses without balancing the rest of the equation on both sides. Since that +1 will be multiplied by a coefficient of 3, you should add 3 to the right side of the equation to match what you've done on the left:
Then you can factor the quadratic on the left into perfect square form, and subtract 3 from both sides to reset to 0:
This provides you with Vertex Form, so you can say that and , making the vertex and the Line of Symmetry just the x-coordinate of .
Example Question #6 : Graph A Quadratic Function Using Intercepts, Vertex, And Axis Of Symmetry
What is the line of symmetry of the parabola formed by ?
x=2
x=4
x=-4
x=1
x=-1
x=-1
The line of symmetry of a parabola is the x-coordinate of its vertex, so you can solve this problem by taking the given quadratic and converting to Vertex Form, , where is the vertex. To do so, focus on the and terms first, pulling them aside and factoring out the common 4 so that you have your coefficient:
Next, think of which Perfect Square quadratic you can form using the terms in parentheses. , so if you add 1 within the parentheses you can treat it as a perfect square to match Vertex Form. Of course, you can't just add 1 -- which will be multiplied by the coefficient of 4 -- without accounting for it on the other side of the equation. So as you transform the parentheses on the left to match Vertex Form and add the +1 within parentheses to do so, also add 4 to the right hand side to stay balanced:
Now you can factor the quadratic to Perfect Square form, and subtract 4 from both sides to finish Vertex Form:
This means that and , so the line of symmetry - which is the x-coordinate - is at .
Example Question #1 : Express A Quadratic Function In Vertex Form
Which of the following is the given quadratic equation in vertex form?
To solve for the vertex form, we must start by completing the square:
Example Question #1 : Express A Quadratic Function In Vertex Form
Which of the following is the appropriate vertex form of the following quadratic equation?
This process outlines how to convert a quadratic function to vertex form:
Example Question #2 : Express A Quadratic Function In Vertex Form
Give the coordinate pair of the vertex of this quadratic function .
None of the other answers.
Expressing quadratic functions in the vertex form is basically just changing the format of the equation to give us different information, namely the vertex. In order for us to change the function into this format we must have it in standard form . After that, our goal is to change the function into the form . We do so as follows:
subtract the constant over to the other side
halve the b term, square it, and add to both sides.
Now factor the left side.
now simplify the right side and move that number back over to the left side and you will be left with . I recommend looking up an example with numbers before you begin or at least recognizing that the fractions will end up being whole numbers in most problems. Below is specific explanation of the problem at hand. Try to use the generic equation to find the answer before following the step by step approach below.
move the constant over
halve the b term and add to both sides
factor the left side and simplify the right
move the constant over to achieve vertex form
is the final answer with vertex at (-1,-7). Note that the formula is .
try this shortcut after you have mastered the steps: . Make sure you recognize that this formula gives you an x and y coordinate for the vertex and that each coordinate of the pair is fraction in the formula. This will give you the vertex of the equation if it is in standard form. However, don't rely on this as completing the square is also a method for finding the roots. So you need to know both methods before you cut the corner.
Example Question #1 : Express A Quadratic Function In Vertex Form
Express the following quadratic equation in vertex form.
To get the equation into vertex form, we factor the largest constant from the terms with a degree of greater than or equal to 1.
We then complete the square by following these steps
- finding half of the coefficients of the term
- squaring that result
- and then adding that square to the expression for .
Keep in mind that what is done on one side of the equation must be done on the other.
And factoring the quadratic polynomial of x
we get
Example Question #1 : Find The Distance Between A Point And A Line
Find the minimum distance between the point and the following line:
The minimum distance from the point to the line would be found by drawing a segment perpendicular to the line directly to the point. Our first step is to find the equation of the new line that connects the point to the line given in the problem. Because we know this new line is perpendicular to the line we're finding the distance to, we know its slope will be the negative inverse of the line its perpendicular to. So if the line we're finding the distance to is:
Then its slope is -1/3, so the slope of a line perpendicular to it would be 3. Now that we know the slope of the line that will give the shortest distance from the point to the given line, we can plug the coordinates of our point into the formula for a line to get the full equation of the new line:
Now that we know the equation of our perpendicular line, our next step is to find the point where it intersects the line given in the problem:
So if the lines intersect at x=0, we plug that value into either equation to find the y coordinate of the point where the lines intersect, which is the point on the line closest to the point given in the problem and therefore tells us the location of the minimum distance from the point to the line:
So we now know we want to find the distance between the following two points:
and
Using the following formula for the distance between two points, which we can see is just an application of the Pythagorean Theorem, we can plug in the values of our two points and calculate the shortest distance between the point and line given in the problem:
Which we can then simplify by factoring the radical:
Example Question #2 : Find The Distance Between A Point And A Line
What is the shortest distance between the line and the origin?
The shortest distance from a point to a line is always going to be along a path perpendicular to that line. To be perpendicular to our line, we need a slope of .
To find the equation of our line, we can simply use point-slope form, using the origin, giving us
which simplifies to .
Now we want to know where this line intersects with our given line. We simply set them equal to each other, giving us .
If we multiply each side by , we get .
We can then add to each side, giving us .
Finally we divide by , giving us .
This is the x-coordinate of their intersection. To find the y-coordinate, we plug into , giving us .
Therefore, our point of intersection must be .
We then use the distance formula using and the origin.
This give us .
Example Question #3 : Find The Distance Between A Point And A Line
Find the distance from point to the line .
Draw a line that connects the point and intersects the line at a perpendicular angle.
The vertical distance from the point to the line will be the difference of the 2 y-values.
The distance can never be negative.
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