Positive Definite Negative Definite


The Discriminant and the Quadratic Graph


The discriminant of the quadratic equation $ax^2+bx+c=0$ is $\Delta = b^2-4ac$.

We used the discriminant to determine the number of real roots of the quadratic equation. If they exist, these roots correspond to $x$-intercepts of the quadratic $y = ax^2+bx+c$. The discriminant tells us about the relationship between a quadratic function and the $x$-axis.

Case 1: $y=x^2-2x+2$

\( \begin{align} \displaystyle
a &= 1, b=-2, c=2 \\
\Delta &= b^2-4ac \\
&= (-2)^2 – 4 \times 1 \times 2 \\
&= -4 \lt 0 \\
\end{align} \)
The graph does not cut the $x$-axis and is always above the $x$-axis, which means $y=x^2-2x+2 \gt 0$.
This case is Positive Definite because $x^2-2x+2$ is always positive for all real $x$.

Case 2: $y=x^2-2x+1$

\( \begin{align} \displaystyle
a &= 1, b=-2, c=1 \\
\Delta &= b^2-4ac \\
&= (-2)^2 – 4 \times 1 \times 1 \\
&= 0 \\
\end{align} \)
The graph touches the $x$-axis the graph is mostly above the $x$-axis, which means $y=x^2-2x+1 \ge 0$.
This case is Not Positive Definite because $x^2-2x+1$ is not always positive for all real $x$.

Case 3: $y=x^2-2x$

\( \begin{align} \displaystyle
a &= 1, b=-2, c=0 \\
\Delta &= b^2-4ac \\
&= (-2)^2 – 4 \times 1 \times 0 \\
&= 4 \gt 0 \\
\end{align} \)
The graph cuts the $x$-axis twice and and the graph is above or below the $x$-axis depending on $x$ values.
This case is Not Positive Definite because $x^2-2x$ is not always positive for all real $x$.

Case 4: $y=-x^2+2x-2$

\( \begin{align} \displaystyle
a &= -1, b=2, c=-2 \\
\Delta &= b^2-4ac \\
&= 2^2 – 4 \times (-1) \times (-2) \\
&= -4 \lt 0 \\
\end{align} \)
The graph does not cut the $x$-axis and is always below the $x$-axis, which means $y=-x^2+2x-2 \lt 0$.
This case is Negative Definite because $-x^2+2x-2$ is always negative for all real $x$.

Case 5: $y=-x^2+2x-1$

\( \begin{align} \displaystyle
a &= -1, b=2, c=-1 \\
\Delta &= b^2-4ac \\
&= 2^2 – 4 \times (-1) \times (-1) \\
&= 0 \\
\end{align} \)
The graph touches the $x$-axis the graph is mostly below the $x$-axis, which means $y=-x^2+2x-1 \le 0$.
This case is Not Positive Definite because $x^2-2x+1$ is not always negative for all real $x$.

Case 6: $y=-x^2+2x$

\( \begin{align} \displaystyle
a &= -1, b=2, c=0 \\
\Delta &= b^2-4ac \\
&= 2^2 – 4 \times (-1) \times 0 \\
&= 4 \gt 0 \\
\end{align} \)
The graph cuts the $x$-axis twice and and the graph is above or below the $x$-axis depending on $x$ values.
This case is Not Negative Definite because $-x^2+2x$ is not always negative for all real $x$.

Positive Definite


Quadratics which are positive for all real values of $x$ like Case 1: $y=x^2-2x+2$
\( \begin{align} \displaystyle
ax^2+bx+c &\gt 0 \\
a &\gt 0 \\
b^2 – 4ac &\lt 0 \\
\end{align} \)

Negative Definite


Quadratics which are negative for all real values of $x$ like Case 4: $y=-x^2+2x-2$
\( \begin{align} \displaystyle
ax^2+bx+c &\lt 0 \\
a &\lt 0 \\
b^2 – 4ac &\lt 0 \\
\end{align} \)

Example 1

Use the discriminant to determine the relationship between the graph of $y=2x^2-4x+5$ and the $x$-axis.

\( \begin{align} \displaystyle
a &= 2, b=-4,c=5 \\
\Delta &= b^2 – 4ac \\
&= (-4)^2 – 4 \times 2 \times 5 \\
&= -24 \lt 0 \\
\end{align} \)
Since $\Delta \lt 0$, the graph does not cut the $x$-axis.
Since $a=2 \gt 0$, the graph is concave up.

The graph is positive definite and lies entirely above the $x$-axis.

Example 2

Use the discriminant to determine the relationship between the graph of $y=-x^2+6x-9$ and the $x$-axis.

\( \begin{align} \displaystyle
a &= -1, b=6,c=-9 \\
\Delta &= b^2 – 4ac \\
&= 6^2 – 4 \times (-1) \times (-9) \\
&= 0 \\
\end{align} \)
Since $\Delta = 0$, the graph touches the $x$-axis.
Since $a=-1 \lt 0$, the graph is concave down.

Example 3

Use the discriminant to determine the relationship between the graph of $y=-x^2+2x-4$ and the $x$-axis.

\( \begin{align} \displaystyle
a &= -1, b=2,c=-4 \\
\Delta &= b^2 – 4ac \\
&= 2^2 – 4 \times (-1) \times (-4) \\
&= -12 \lt 0 \\
\end{align} \)
Since $\Delta \lt 0$, the graph does not cut the $x$-axis.
Since $a=-1 \lt 0$, the graph is concave down.

The graph is negative definite and lies entirely below the $x$-axis.

Example 4

Use the discriminant to determine the relationship between the graph of $y=x^2+x-2$ and the $x$-axis.

\( \begin{align} \displaystyle
a &= 1, b=2,c=-2 \\
\Delta &= b^2 – 4ac \\
&= 2^2 – 4 \times 1 \times (-2) \\
&= 12 \gt 0 \\
\end{align} \)
Since $\Delta \gt 0$, the graph cuts the $x$-axis twice.
Since $a=1 \gt 0$, the graph is concave up.

Example 5

Use the discriminant to determine the relationship between the graph of $y=4x^2-4x+1$ and the $x$-axis.

\( \begin{align} \displaystyle
a &= 4, b=-4,c=1 \\
\Delta &= b^2 – 4ac \\
&= (-4)^2 – 4 \times 4 \times 1 \\
&= 0 \\
\end{align} \)
Since $\Delta = 0$, the graph touches the $x$-axis.
Since $a=4 \gt 0$, the graph is concave up.


Example 6

Use the discriminant to determine the relationship between the graph of $y=-2x^2+3x+1$ and the $x$-axis.

\( \begin{align} \displaystyle
a &= -2, b=3,c=1 \\
\Delta &= b^2 – 4ac \\
&= 3^2 – 4 \times (-2) \times 1 \\
&= 17 \gt 0 \\
\end{align} \)
Since $\Delta \gt 0$, the graph cuts the $x$-axis twice.
Since $a=-2 \lt 0$, the graph is concave down.


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