Find the graph of the following quadratic functions.
f\left(x\right)=x^2-5x+6\\
f\left(x\right)=x^2+2x
The sign of the coefficient of x^2 is positive, therefore the graph opens up (smiles).
The roots of f are:
x^2+2x=0\\x\left(x+2\right) =0
Thus:
- f\left(-2\right) = 0
- f\left(0\right)=0
The only graph that satisfies these relations is:
f\left(x\right)=-2x^2-1
The sign of the coefficient of x^2 is negative, therefore the graph opens down (frowns). Only two graphs satisfy this property.
Furthermore:
f\left(0\right) = -1
One of them can be eliminated. The following is the only acceptable graph:
f\left(x\right)=-x^2-2x -\dfrac{3}{4}
The sign of the coefficient of x^2 is negative, therefore the graph opens down (frowns). Since three out of four graphs are very similar, we need to find the vertex of the graph to identify it. Recall that If the quadratic is written in the form:
y = a\left(x – h\right)^2 + k,
Then the vertex is the point \left(h,k\right) . Completing the square we have:
f\left(x\right)=-x^2-2x -\dfrac{3}{4} \\= -x^2-2x-1 + \dfrac{1}{4} \\= -\left(x+1\right)^2 + \dfrac{1}{4}
The vertex of the graph is:
\left(-1,\dfrac{1}{4}\right)
Therefore, the graph of f is:
f\left(x\right)=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}
The sign of the coefficient of x^2 is positive, therefore the graph opens up (smiles). Recall that the quadratic is written in the form:
y = a\left(x – h\right)^2 + k,
Then the vertex is the point \left(h,k\right) . The vertex of the parabola is:
\left(\dfrac{1}{2},\dfrac{1}{2}\right)
Therefore, the graph of f is:
f\left(x\right)=x^2-x-1
The sign of the coefficient of x^2 is negative, therefore the graph opens down (frowns). Since three out of four graphs are very similar, we need to find the vertex of the graph to identify it. Recall that If the quadratic is written in the form:
y = a\left(x – h\right)^2 + k,
Then the vertex is the point \left(h,k\right) . Completing the square we have:
f\left(x\right)=x^2-x-1 \\= x^2-x+ \dfrac{1}{4} - \dfrac{5}{4}\\= \left(x-\dfrac{1}{2}\right)^2 - \dfrac{5}{4}
The vertex of the graph is:
\left(\dfrac{1}{2},-\dfrac{5}{4}\right)
Therefore, the graph of f is:
f\left(x\right)=2x^2+4x+2
The sign of the coefficient of x^2 is positive, therefore the graph opens up (smiles). Recall that If the quadratic is written in the form:
y = a\left(x – h\right)^2 + k,
Then the vertex is the point \left(h,k\right) . Completing the square we have:
f\left(x\right)=2x^2+4x+2 \\= 2\left(x^2+2x+1\right)\\= 2\left(x+1\right)^2
The vertex of the graph is:
\left(-1{,}0\right)
Since f\left(0\right)=2, the graph of f is:
We have:
- f\left(2\right) = 0
- f\left(0\right)=6
The only graph that satisfies these relations is the following:
