Find the vertex, focus, and directrix of the parabola with the gi...

Question

Find the vertex, focus, and directrix of the parabola with the given equation. Then graph the parabola. x^2 - 4x - 2y = 0

Accepted Answer

Hello Today we're going to be using the given equation to identify the graph of the parabola. So what we are given is X squared minus 10 X plus four. Y is equal to 29. So we're going to need to rewrite this equation into the standard form of a parabola. We can do this by grouping up the X terms together and moving the Y term to the right hand side of the equation. Doing this is going to give us x squared minus 10. X is equal to negative four Y plus 29. And in order to simplify the left hand side we can go ahead and do completing the square on the X group, completing the square states that we take the middle term which is negative 10 in this case divided by two and then square that quantity and that's going to give us the value of 25. So we're going to be adding 25 to both the left and right hand side of the equation and that's going to give us X squared minus 10, X plus 25 is equal to negative four, Y plus 29 plus 25. In order to balance the equation. Now we just need to go ahead and simplify the left hand side is a fact herbal try no meal as X squared minus 10 X plus factors to the quantity x minus five squared and on the right hand side, 29 plus 25 is going to simplify to give us 54. So we have negative four Y plus 54. Now we want this y value to be positive in order to do that. We need to factor out a negative four from the right hand side of the equation. And doing that is going to give us x minus five squared is equal to negative four times Y plus 54 divided by negative four, which is going to simplify to give us 27/2. And so what we have is the standard form of the equation of a parabola not located at the origin. So the standard form of this equation is given to us as x minus h squared is equal to four, p times y minus case. And here this is the equation of a parabola with respect to the y axis since the x quantity is squared and since the leading coefficient is negative, this is going to be a parabola that opens downward and not located at the origin. So before we figure out anything else, let's go ahead and identify the vertex of this parabola. The vertex is going to be given to us in the form of h comma K. And we can get our H and K values by taking a look at the X and y quantities. See here we have x minus five for the X quantity and this is in the standard form of x minus H. This is where H is going to equal to five and for the y quantity we have Y minus 27/2. And this is going to be where K is equal to 27/2. With that being said, the vertex of the parabola is going to be located at five comma 27/2. So now that we have the vertex, we need to go ahead and identify the fosse Or at least we can identify the distance of the fosse from the Vertex C. The focal constant is given to us as four p. And in our current equation the focal constant is equal to -4. So if we take four p and set it equal to negative four and solve, we're going to get the distance of the fosse from the vertex. If we divide both sides of this equation by four, we get P is equal to negative one. This means that the photo site is going to be located one unit below the vertex. So we're going to take negative one and subtract it from the y value of the vertex, which in this case is 27/2. So we have five comma 27/2 minus one. And that's going to simplify to give us five comma 25/2, which is going to be the location of the fosse. Now we need to figure out the location of the directory X. Now the direct tricks is going to be a horizontal line which is given to us as why is equal to some constant. So if the fosse is one unit below the vertex, that means that the direct tricks is going to be one unit above the vertex. The current vertex has a Y value of 27/2. And if the directory is located one unit above this, we're going to be adding one to this value. And that means that the direct tricks is going to be represented as y is equal to 29/2. Now that we have the vertex fosse and direct tricks, we just need to select the graph that accurately represents this information and that graph is going to be C. See here the vertex is five comma 27/2, which we've identified up above The focus is given to us as five comma 25/2 which is slightly below the current vertex. And the direct tricks is given to us as y is equal to 29/2, which is slightly above the current vertex. So with that being said, the answer to this problem is going to be C. So I hope this video helps you in understanding how to identify the graph of a parabola. And I'll go ahead and see you all in the next video

Find the vertex, focus, and directrix of the parabola with the given equation. Then graph the parabola. x^2 - 4x - 2y = 0

Key Concepts

Parabola Properties

Completing the Square

Graphing Quadratic Functions

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