Find A Quadratic Function

mathdad

mathdad

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Find a quadratic function whose x-intercepts -4 and 2 and whose range is [-18, infinity).

How is this done? The textbook does not have a sample question for this problem. I seek the needed steps leading to the quadratic function.
 
Another approach is to use the factored form a(x-p)(x-q) = 0, knowing the two x-intercepts, and then solve for a using the known minimum. (Or maybe this is just the second step after MarkFL's idea.)
 
MarkFL said:
We know the axis of symmetry will be midway between the roots, and so where must the vertex be?
Click to expand...

Mark,

I had no idea that you are also here. Good to hear from you. I know the axis of symmetry is found by using x = -b/2a. Are you saying to add the given roots and divide by 2?
 
Dr.Peterson said:
Another approach is to use the factored form a(x-p)(x-q) = 0, knowing the two x-intercepts, and then solve for a using the known minimum. (Or maybe this is just the second step after MarkFL's idea.)
Click to expand...

I will work on it and post my reply when time allows.
 
mathdad said:
Mark,

I had no idea that you are also here. Good to hear from you. I know the axis of symmetry is found by using x = -b/2a. Are you saying to add the given roots and divide by 2?
Click to expand...

Yeah, I get around...:)

Yes,, you're essentially using the mid-point formula. So, where would the vertex be?
 
MarkFL said:
Yeah, I get around...:)

Yes,, you're essentially using the mid-point formula. So, where would the vertex be?
Click to expand...

I have an idea about how to do this problem. I will post my work when time allows.
 
MarkFL said:
We know the axis of symmetry will be midway between the roots, and so where must the vertex be?
Click to expand...

This parabola opens upward.

The general equation for an up-opening parabola is:

y = a(x-h)^2 + k with a > 0 and its vertex the point (h, k).

If the roots are -4 and 2, we can express this as (-4, 0) and (2, 0).

Let M = midpoint = vertex point.

M = [(-4 + 2)/2, (0 + 0)/2)]

M = (-1, 0).

Is this ok so far?
 
mathdad said:
This parabola opens upward.

The general equation for an up-opening parabola is:

y = a(x-h)^2 + k with a > 0 and its vertex the point (h, k).

If the roots are -4 and 2, we can express this as (-4, 0) and (2, 0).

Let M = midpoint = vertex point.

M = [(-4 + 2)/2, (0 + 0)/2)]

M = (-1, 0).

Is this ok so far?
Click to expand...

Yes, this means the axis of symmetry is the line \(x=-1\)...so, based on the given range, where is the vertex?
 
mathdad said:
... Let M ... = vertex point

... M = (-1, 0).

Is this ok so far?
Click to expand...
Hi. Your coordinates for M are half correct.

The vertex x-coordinate is the midpoint between the intercepts, and you found it, but the y-coordinate is not 0.

As the vertex is the lowest point on a parabola that opens upward, its y-coordinate is the smallest value in the function's range. They gave you the range.

?
 
Otis said:
Hi. Your coordinates for M are half correct.

The vertex x-coordinate is the midpoint between the intercepts, and you found it, but the y-coordinate is not 0.

As the vertex is the lowest point on a parabola that opens upward, its y-coordinate is the smallest value in the function's range. They gave you the range.

?
Click to expand...

Vertex = (-1, -18).
 
Yes the vertex is (-1 -18).
So, now using the intercept form y=a(x-p)(x-q), you have enough info to work out "a". Or you could use vertex form also.
 
Harry_the_cat said:
Yes the vertex is (-1 -18).
So, now using the intercept form y=a(x-p)(x-q), you have enough info to work out "a". Or you could use vertex form also.
Click to expand...

In that form you provided, p = -1, q = -18 but what does x equal to?
 
No, in intercept form y = a (x-p)(x-q), p and q are the intercepts. That's why it is called intercept form.

In vertex form, the h and k represent the x and y coordinates of the vertex respectively.
 
Harry_the_cat said:
No, in intercept form y = a (x-p)(x-q), p and q are the intercepts. That's why it is called intercept form.

In vertex form, the h and k represent the x and y coordinates of the vertex respectively.
Click to expand...

-18 = a (-1-(-4))(-1-2)

-18 = a(-1 + 4)(-3)

-18 = a(3)(-3)

-18 = -9a

-18/-9 = a

2 = a

y = 2(x - h)^2 + k

y = 2(x -(-1))^2 + (-18)

y = 2(x + 1)^2 - 18
 
or
y = 2(x+4)(x-2) ….. intercept form
or
y = 2x^2 + 4x -16 …. standard form

They are all the same equation just in different forms.
 
mathdad said:
... I know the axis of symmetry is found by using x = -b/(2a) ...
Click to expand...
Hi. Here are a few closing comments.

Note the grouping symbols that I inserted around the denominator above (shown in red). When we type ratios that contain more than a single number (within a numerator or a denominator), we need to enclose the numerator or denominator in grouping symbols, to avoid ambiguity.

For example, people (and software) may interpret an expression like -b/2a as the fraction -b/2 times a. The grouping symbols above make clear that a appears in the denominator, instead.

You're correct that the expression -b/(2a) gives the x-coordinate of the vertex, but we were not provided values for coefficients a,b,c in the quadratic polynomial. Now you have another way. Given the intercepts of a parabola (that opens upward or downward, not sideways), the midpoint between those intercepts is the x-coordinate of the vertex. Both facts about the horizontal location of the vertex are good to remember.

My last comment concerns intercept form versus factored form. They're both the same, but the intercepts are shown explicity only when the factors are written in terms of subtraction.

Here's an example, using the quadratic polynomial x^2+5x+6 whose x-intercepts are -2 and -3. Both factorizations below are correct, but only the second form shows the x-intercepts explicitly.

y = (x + 2)(x + 3)

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

In other words, if you're given (x+2)(x+3) and you're told that it's intercept form, then it is up to you to remember to view the factors in terms of subtraction (in order to see the intercepts explicitly).

In this exercise, we have:

y = 2(x + 4)(x - 2)

Can you see from this factorization that the x-intercepts are -4 and 2? (The x-intercepts are the numbers being subtracted from x.)

?
 
Otis said:
Hi. Here are a few closing comments.

Note the grouping symbols that I inserted around the denominator above (shown in red). When we type ratios that contain more than a single number (within a numerator or a denominator), we need to enclose the numerator or denominator in grouping symbols, to avoid ambiguity.

For example, people (and software) may interpret an expression like -b/2a as the fraction -b/2 times a. Ther grouping symbols above make clear that a appears in the denominator, instead.

You're correct that the expression -b/(2a) gives the x-coordinate of the vertex, but we were not provided values for coefficients a,b,c in the quadratic polynomial. Now you have another way. Given the intercepts of a parabola (that opens upward or downward, not sideways), the midpoint between those intercepts is the x-coordinate of the vertex. Both facts about the horizontal location of the vertex are good to remember.

My last comment concerns intercept form versus factored form. They both look similar, but the intercepts are shown only when the factors containing x are written in terms of subtraction.

Here's an example, using the quadratic polynomial x^2+5x+6 whose x-intercepts are -2 and -3. Both factorizations below are correct, but only the second form shows the x-intercepts explicitly.

y = (x + 2)(x + 3)

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

In other words, if you're given (x+2)(x+3) and you're told that it's intercept form, then it is up to you to remember to rewrite the factors in terms of subtraction (to see the intercepts).

In this exercise, we have:

y = 2(x + 4)(x - 2)

Can you see from this factorization that the x-intercepts are -4 and 2? (The x-intercepts are the numbers being subtracted from x.)

?
Click to expand...

You said:

y = 2(x + 4)(x - 2)

Can you see from this factorization that the x-intercepts are -4 and 2? (The x-intercepts are the numbers being subtracted from x.)

Yes, is my answer. If I set each factor to 0,
x = -4 and x = 2.
 
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