Discriminant when graph lies above or below the x axis.
$begingroup$
Suppose a quadratic equation has been given where the a value (ax^2 + bx + c) is a positive and it has been said that the graph of the equation lies above the x-axis- what is the discriminant?
For example- 2x^2 + kx - 5 = y; the graph lies above the x-axis- find the possible values of k. I know how to solve it, just not sure what discriminant to take.
In the same sense, if the a value is a negative, and the graph is said to be lying above the x axis, is the discriminant > 0 ?
quadratics discriminant
asked Dec 9 '18 at 14:33
De SithDe Sith
1
$endgroup$
add a comment |
$begingroup$
Suppose a quadratic equation has been given where the a value (ax^2 + bx + c) is a positive and it has been said that the graph of the equation lies above the x-axis- what is the discriminant?
For example- 2x^2 + kx - 5 = y; the graph lies above the x-axis- find the possible values of k. I know how to solve it, just not sure what discriminant to take.
In the same sense, if the a value is a negative, and the graph is said to be lying above the x axis, is the discriminant > 0 ?
quadratics discriminant
asked Dec 9 '18 at 14:33
De SithDe Sith
1
$endgroup$
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:36
add a comment |
$begingroup$
Suppose a quadratic equation has been given where the a value (ax^2 + bx + c) is a positive and it has been said that the graph of the equation lies above the x-axis- what is the discriminant?
For example- 2x^2 + kx - 5 = y; the graph lies above the x-axis- find the possible values of k. I know how to solve it, just not sure what discriminant to take.
In the same sense, if the a value is a negative, and the graph is said to be lying above the x axis, is the discriminant > 0 ?
quadratics discriminant
asked Dec 9 '18 at 14:33
De SithDe Sith
1
$endgroup$
Suppose a quadratic equation has been given where the a value (ax^2 + bx + c) is a positive and it has been said that the graph of the equation lies above the x-axis- what is the discriminant?
For example- 2x^2 + kx - 5 = y; the graph lies above the x-axis- find the possible values of k. I know how to solve it, just not sure what discriminant to take.
In the same sense, if the a value is a negative, and the graph is said to be lying above the x axis, is the discriminant > 0 ?
quadratics discriminant
quadratics discriminant
asked Dec 9 '18 at 14:33
De SithDe Sith
1
asked Dec 9 '18 at 14:33
De SithDe Sith
1
asked Dec 9 '18 at 14:33
De SithDe Sith
1
asked Dec 9 '18 at 14:33
De SithDe Sith
1
asked Dec 9 '18 at 14:33
De SithDe Sith
1
1
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:36
add a comment |
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:36
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:36
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:36
add a comment |
1 Answer
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$begingroup$
You need the discrimination to be negative so that the quadratic function has no real roots.
$$Delta < 0 iff b^2-4ac < 0$$
Therefore, the graph either lies entirely above or below the $x$-axis. This is determined by the value of $a$.
If $a > 0$, then the graph is concave upward and has a minimum. Clearly, no roots and a minimum implies the graph lies entirely above the $x$-axis.
If $a < 0$, then the graph is concave downward and has a maximum. Clearly, no roots and a maximum implies the graph lies entirely below the $x$-axis.
For your example, you have $y = 2x^2+kx-5$, so you use
$$k^2-4(2)(-5) < 0 iff k^2+40 < 0 iff k^2 < -40$$
Clearly, the inequality is not true for real values of $k$, since $k^2 geq 0$, so no value of $k$ allows the entire graph to remain above the $x$-axis.
answered Dec 9 '18 at 14:37
KM101KM101
5,9231524
$endgroup$
$begingroup$
but if they have said that the graph lies above the x-axis- and is a downward concave (a<0), it means that the maximum point is above the x axis right? So if is a downward concave, wont the graph intersect the x axis at two points, making the discriinant greater than 0?
$endgroup$
– De Sith
Dec 9 '18 at 14:41
$begingroup$
A concave downward graph can’t lie entirely above the $x$-axis. Similarly, a concave upward graph can’t lie entirely below the $x$-axis. It’s pretty obvious. In other words, a question like that would be pointless. Here, we’re talking about a concave upward graph above the $x$-axis and a concave downward graph below the $x$-axis. (Then, you let $Delta < 0$.)
$endgroup$
– KM101
Dec 9 '18 at 14:45
$begingroup$
yeah, so when they say the graph is above the x axis, it is not entirely above the x axis, I understand that. If the 'a' value is a negative, for a graph that the question says lies above the x axis- the maximum point has to be above the x axis right? Otherwise it would not be above the x axis at all. So if the maximum is above the x axis- won't the discriminant be > 0 since the graph will cut the x-axis at two x values?
$endgroup$
– De Sith
Dec 9 '18 at 14:50
$begingroup$
Yes, the discriminant would be positive then, since there are two real solutions.
$endgroup$
– KM101
Dec 9 '18 at 14:52
add a comment |
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$begingroup$
You need the discrimination to be negative so that the quadratic function has no real roots.
$$Delta < 0 iff b^2-4ac < 0$$
Therefore, the graph either lies entirely above or below the $x$-axis. This is determined by the value of $a$.
If $a > 0$, then the graph is concave upward and has a minimum. Clearly, no roots and a minimum implies the graph lies entirely above the $x$-axis.
If $a < 0$, then the graph is concave downward and has a maximum. Clearly, no roots and a maximum implies the graph lies entirely below the $x$-axis.
For your example, you have $y = 2x^2+kx-5$, so you use
$$k^2-4(2)(-5) < 0 iff k^2+40 < 0 iff k^2 < -40$$
Clearly, the inequality is not true for real values of $k$, since $k^2 geq 0$, so no value of $k$ allows the entire graph to remain above the $x$-axis.
answered Dec 9 '18 at 14:37
KM101KM101
5,9231524
$endgroup$
$begingroup$
but if they have said that the graph lies above the x-axis- and is a downward concave (a<0), it means that the maximum point is above the x axis right? So if is a downward concave, wont the graph intersect the x axis at two points, making the discriinant greater than 0?
$endgroup$
– De Sith
Dec 9 '18 at 14:41
$begingroup$
A concave downward graph can’t lie entirely above the $x$-axis. Similarly, a concave upward graph can’t lie entirely below the $x$-axis. It’s pretty obvious. In other words, a question like that would be pointless. Here, we’re talking about a concave upward graph above the $x$-axis and a concave downward graph below the $x$-axis. (Then, you let $Delta < 0$.)
$endgroup$
– KM101
Dec 9 '18 at 14:45
$begingroup$
yeah, so when they say the graph is above the x axis, it is not entirely above the x axis, I understand that. If the 'a' value is a negative, for a graph that the question says lies above the x axis- the maximum point has to be above the x axis right? Otherwise it would not be above the x axis at all. So if the maximum is above the x axis- won't the discriminant be > 0 since the graph will cut the x-axis at two x values?
$endgroup$
– De Sith
Dec 9 '18 at 14:50
$begingroup$
Yes, the discriminant would be positive then, since there are two real solutions.
$endgroup$
– KM101
Dec 9 '18 at 14:52
add a comment |
$begingroup$
You need the discrimination to be negative so that the quadratic function has no real roots.
$$Delta < 0 iff b^2-4ac < 0$$
Therefore, the graph either lies entirely above or below the $x$-axis. This is determined by the value of $a$.
If $a > 0$, then the graph is concave upward and has a minimum. Clearly, no roots and a minimum implies the graph lies entirely above the $x$-axis.
If $a < 0$, then the graph is concave downward and has a maximum. Clearly, no roots and a maximum implies the graph lies entirely below the $x$-axis.
For your example, you have $y = 2x^2+kx-5$, so you use
$$k^2-4(2)(-5) < 0 iff k^2+40 < 0 iff k^2 < -40$$
Clearly, the inequality is not true for real values of $k$, since $k^2 geq 0$, so no value of $k$ allows the entire graph to remain above the $x$-axis.
answered Dec 9 '18 at 14:37
KM101KM101
5,9231524
$endgroup$
$begingroup$
but if they have said that the graph lies above the x-axis- and is a downward concave (a<0), it means that the maximum point is above the x axis right? So if is a downward concave, wont the graph intersect the x axis at two points, making the discriinant greater than 0?
$endgroup$
– De Sith
Dec 9 '18 at 14:41
$begingroup$
A concave downward graph can’t lie entirely above the $x$-axis. Similarly, a concave upward graph can’t lie entirely below the $x$-axis. It’s pretty obvious. In other words, a question like that would be pointless. Here, we’re talking about a concave upward graph above the $x$-axis and a concave downward graph below the $x$-axis. (Then, you let $Delta < 0$.)
$endgroup$
– KM101
Dec 9 '18 at 14:45
$begingroup$
yeah, so when they say the graph is above the x axis, it is not entirely above the x axis, I understand that. If the 'a' value is a negative, for a graph that the question says lies above the x axis- the maximum point has to be above the x axis right? Otherwise it would not be above the x axis at all. So if the maximum is above the x axis- won't the discriminant be > 0 since the graph will cut the x-axis at two x values?
$endgroup$
– De Sith
Dec 9 '18 at 14:50
$begingroup$
Yes, the discriminant would be positive then, since there are two real solutions.
$endgroup$
– KM101
Dec 9 '18 at 14:52
add a comment |
$begingroup$
You need the discrimination to be negative so that the quadratic function has no real roots.
$$Delta < 0 iff b^2-4ac < 0$$
Therefore, the graph either lies entirely above or below the $x$-axis. This is determined by the value of $a$.
If $a > 0$, then the graph is concave upward and has a minimum. Clearly, no roots and a minimum implies the graph lies entirely above the $x$-axis.
If $a < 0$, then the graph is concave downward and has a maximum. Clearly, no roots and a maximum implies the graph lies entirely below the $x$-axis.
For your example, you have $y = 2x^2+kx-5$, so you use
$$k^2-4(2)(-5) < 0 iff k^2+40 < 0 iff k^2 < -40$$
Clearly, the inequality is not true for real values of $k$, since $k^2 geq 0$, so no value of $k$ allows the entire graph to remain above the $x$-axis.
answered Dec 9 '18 at 14:37
KM101KM101
5,9231524
$endgroup$
You need the discrimination to be negative so that the quadratic function has no real roots.
$$Delta < 0 iff b^2-4ac < 0$$
Therefore, the graph either lies entirely above or below the $x$-axis. This is determined by the value of $a$.
If $a > 0$, then the graph is concave upward and has a minimum. Clearly, no roots and a minimum implies the graph lies entirely above the $x$-axis.
If $a < 0$, then the graph is concave downward and has a maximum. Clearly, no roots and a maximum implies the graph lies entirely below the $x$-axis.
For your example, you have $y = 2x^2+kx-5$, so you use
$$k^2-4(2)(-5) < 0 iff k^2+40 < 0 iff k^2 < -40$$
Clearly, the inequality is not true for real values of $k$, since $k^2 geq 0$, so no value of $k$ allows the entire graph to remain above the $x$-axis.
answered Dec 9 '18 at 14:37
KM101KM101
5,9231524
edited Dec 9 '18 at 14:41
answered Dec 9 '18 at 14:37
KM101KM101
5,9231524
answered Dec 9 '18 at 14:37
KM101KM101
5,9231524
answered Dec 9 '18 at 14:37
KM101KM101
5,9231524
5,9231524
$begingroup$
but if they have said that the graph lies above the x-axis- and is a downward concave (a<0), it means that the maximum point is above the x axis right? So if is a downward concave, wont the graph intersect the x axis at two points, making the discriinant greater than 0?
$endgroup$
– De Sith
Dec 9 '18 at 14:41
$begingroup$
A concave downward graph can’t lie entirely above the $x$-axis. Similarly, a concave upward graph can’t lie entirely below the $x$-axis. It’s pretty obvious. In other words, a question like that would be pointless. Here, we’re talking about a concave upward graph above the $x$-axis and a concave downward graph below the $x$-axis. (Then, you let $Delta < 0$.)
$endgroup$
– KM101
Dec 9 '18 at 14:45
$begingroup$
yeah, so when they say the graph is above the x axis, it is not entirely above the x axis, I understand that. If the 'a' value is a negative, for a graph that the question says lies above the x axis- the maximum point has to be above the x axis right? Otherwise it would not be above the x axis at all. So if the maximum is above the x axis- won't the discriminant be > 0 since the graph will cut the x-axis at two x values?
$endgroup$
– De Sith
Dec 9 '18 at 14:50
$begingroup$
Yes, the discriminant would be positive then, since there are two real solutions.
$endgroup$
– KM101
Dec 9 '18 at 14:52
add a comment |
$begingroup$
but if they have said that the graph lies above the x-axis- and is a downward concave (a<0), it means that the maximum point is above the x axis right? So if is a downward concave, wont the graph intersect the x axis at two points, making the discriinant greater than 0?
$endgroup$
– De Sith
Dec 9 '18 at 14:41
$begingroup$
A concave downward graph can’t lie entirely above the $x$-axis. Similarly, a concave upward graph can’t lie entirely below the $x$-axis. It’s pretty obvious. In other words, a question like that would be pointless. Here, we’re talking about a concave upward graph above the $x$-axis and a concave downward graph below the $x$-axis. (Then, you let $Delta < 0$.)
$endgroup$
– KM101
Dec 9 '18 at 14:45
$begingroup$
yeah, so when they say the graph is above the x axis, it is not entirely above the x axis, I understand that. If the 'a' value is a negative, for a graph that the question says lies above the x axis- the maximum point has to be above the x axis right? Otherwise it would not be above the x axis at all. So if the maximum is above the x axis- won't the discriminant be > 0 since the graph will cut the x-axis at two x values?
$endgroup$
– De Sith
Dec 9 '18 at 14:50
$begingroup$
Yes, the discriminant would be positive then, since there are two real solutions.
$endgroup$
– KM101
Dec 9 '18 at 14:52
$begingroup$
but if they have said that the graph lies above the x-axis- and is a downward concave (a<0), it means that the maximum point is above the x axis right? So if is a downward concave, wont the graph intersect the x axis at two points, making the discriinant greater than 0?
$endgroup$
– De Sith
Dec 9 '18 at 14:41
$begingroup$
but if they have said that the graph lies above the x-axis- and is a downward concave (a<0), it means that the maximum point is above the x axis right? So if is a downward concave, wont the graph intersect the x axis at two points, making the discriinant greater than 0?
$endgroup$
– De Sith
Dec 9 '18 at 14:41
$begingroup$
A concave downward graph can’t lie entirely above the $x$-axis. Similarly, a concave upward graph can’t lie entirely below the $x$-axis. It’s pretty obvious. In other words, a question like that would be pointless. Here, we’re talking about a concave upward graph above the $x$-axis and a concave downward graph below the $x$-axis. (Then, you let $Delta < 0$.)
$endgroup$
– KM101
Dec 9 '18 at 14:45
$begingroup$
A concave downward graph can’t lie entirely above the $x$-axis. Similarly, a concave upward graph can’t lie entirely below the $x$-axis. It’s pretty obvious. In other words, a question like that would be pointless. Here, we’re talking about a concave upward graph above the $x$-axis and a concave downward graph below the $x$-axis. (Then, you let $Delta < 0$.)
$endgroup$
– KM101
Dec 9 '18 at 14:45
$begingroup$
yeah, so when they say the graph is above the x axis, it is not entirely above the x axis, I understand that. If the 'a' value is a negative, for a graph that the question says lies above the x axis- the maximum point has to be above the x axis right? Otherwise it would not be above the x axis at all. So if the maximum is above the x axis- won't the discriminant be > 0 since the graph will cut the x-axis at two x values?
$endgroup$
– De Sith
Dec 9 '18 at 14:50
$begingroup$
yeah, so when they say the graph is above the x axis, it is not entirely above the x axis, I understand that. If the 'a' value is a negative, for a graph that the question says lies above the x axis- the maximum point has to be above the x axis right? Otherwise it would not be above the x axis at all. So if the maximum is above the x axis- won't the discriminant be > 0 since the graph will cut the x-axis at two x values?
$endgroup$
– De Sith
Dec 9 '18 at 14:50
$begingroup$
Yes, the discriminant would be positive then, since there are two real solutions.
$endgroup$
– KM101
Dec 9 '18 at 14:52
$begingroup$
Yes, the discriminant would be positive then, since there are two real solutions.
$endgroup$
– KM101
Dec 9 '18 at 14:52
add a comment |
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$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:36