SOLUTION: I need some help on this im just not getting it at all. Find the x-intercepts of the parabola with vertex (3,-2) and y- intercept (0,7). Write your answer in this form: (x1,y1)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: I need some help on this im just not getting it at all. Find the x-intercepts of the parabola with vertex (3,-2) and y- intercept (0,7). Write your answer in this form: (x1,y1)      Log On


   



Question 611676: I need some help on this im just not getting it at all.
Find the x-intercepts of the parabola with vertex (3,-2) and y- intercept (0,7). Write your answer in this form: (x1,y1),(x2,y2). If necessary, round to the nearest hundredth.
Please and thanks
John

Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


In order to find the -intercepts, you need to write the specific function that describes a parabola that has the specific criteria given. In order to create a unique function whose graph is a parabola you need three points. The bad news is, you were only given two points, the vertex and the -intercept. But the good news is, one of the points given is the vertex and parabolas are symmetric about a line passing through the vertex. In this case I'm going to assume that the axis of symmetry is a vertical line. If that is not the case, then 1) this problem is at least one order of magnitude more difficult than the one I'm going to solve, and 2) the problem cannot be solved with only the information given.

Assuming a vertical line of symmetry, namely , there must be a point on the graph with a function value equal to the -coordinate of the -intercept at a horizontal distance from the vertex equal to the horizontal distance of the vertex from the -axis.

The -axis is 3 units horizontally distant from the vertex at (3,-2). Therefore there must be another point on the graph of the parabola 3 units on the OTHER side of the vertex, namely at (6,7).

From this information we can create the quadratic function which describes the set of ordered pairs that is the graph of the desired parabola.

The standard form of a quadratic function is:



substituting the values of the coordinates of any point in for and in the function results in a true statement if and only if the point selected is on the graph of the function.

Hence, if the point (0,7) is on the graph then



which is to say



must be a true statement.

Likewise, and I'll leave verification of the arithmetic to you:



and



must also be true statements.

Using the reduced form of the first result of this analysis we can reduce the 3X3 system of equations to a 2X2 system:





And I'll leave it as an exercise for you to determine that the solution to the original 3X3 system is the ordered triple

resulting in the quadratic function:



From here, set the function equal to 0 (the -intercepts are the two places where the value of the function is 0), and solve using ordinary means. Don't spend much time trying to factor this since it does not factor over the rational numbers. I recommend the quadratic formula.

John

My calculator said it, I believe it, that settles it
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Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
I need some help on this im just not getting it at all.
Find the x-intercepts of the parabola with vertex (3,-2) and y- intercept (0,7). Write your answer in this form: (x1,y1),(x2,y2). If necessary, round to the nearest hundredth.
Please and thanks
John
The graph has the equation

y = a(x - h)² + k  where the vertex = (h,k)

So it has the equation:

y = a(x - 3)² - 2

But we don't know "a" yet.

Draw the two points



Since the vertex is (3,-2), we draw the axis of symmetry
as a vertical line through the vertex:



We can draw the graph approximately:



So you see the x-intercepts are somewhere around 1 and a half and
4 and a half.  But that won't do.  We have to get them to the nearest
hundredth.

Since the graph passes through (0,7), the y-intercept, we substitute
that point in:

y = a(x - 3)² - 2

7 = a(0 - 3)² - 2

7 = a(-3)² - 2

7 = a(9) - 2

7 = 9a - 2

9 = 9a

1 = a

So the equation is

y = 1(x - 3)² - 2

y = (x - 3)² - 2

y = (x - 3)(x - 3) - 2

y = x² - 6x + 9 - 2

y = x² - 6x + 7

To find the x-intercepts, substitute 0 for y

0 = x² - 6x + 7

x² - 6x + 7 = 0

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x+=+%28-%28-6%29+%2B-+sqrt%28+%28-6%29%5E2-4%2A%281%29%2A%287%29+%29%29%2F%282%2A%281%29%29+

x+=+%286+%2B-+sqrt%2836-28+%29%29%2F2+

x+=+%286+%2B-+sqrt%288%29%29%2F2+

x+=+%286+%2B-+sqrt%284%2A2%29%29%2F2+

x+=+%286+%2B-+2sqrt%282%29%29%2F2+

x+=+%282%283+%2B-+sqrt%282%29%29%29%2F2+

x+=+%28cross%282%29%283+%2B-+sqrt%282%29%29%29%2Fcross%282%29+

x+=+3+%2B-+sqrt%282%29+

Using the -, 1.585786438

Using the +, 4.414213562

To the nearest hundredth, the solutions are 1.59 and 4.41

So those two points where the parabola crosses the x-axis

are  (1.59,0) and (4.41,0).

So you write your answer (1.59,0), (4.41,0).

Edwin