Question

I need help with this question please. (It’s also just a homework practice).

Answer 1

We need to graph the next function:

`y=x^2-6x-4`

This is a parabola, and we need 3 points to graph it.

The x-coordinate of a parabola is computed as follows:

`x_V=(-b)/(2a)`

where a and b are the coefficients of the parabola. Substituting with a = 1, and b = -6, we get:

`\begin{gathered} x_V=(-(-6))/(2\cdot1) \\ x_V=3 \end{gathered}`

The y-coordinate of the vertex is found substituting xV into the equation as follows:

`\begin{gathered} y_V=x^2_V-6x_V-4 \\ y_V=3^2-6\cdot3-4 \\ y_V=9-18-4 \\ y_V=-13 \end{gathered}`

The vertex is the point (3, -13)

Substituting x = 2 into the equation of the parabola, we get:

`\begin{gathered} y=2^2-6\cdot2-4 \\ y=4-12-4 \\ y=-12 \end{gathered}`

Substituting x = 4 into the equation of the parabola, we get:

`\begin{gathered} y=4^2-6\cdot4-4 \\ y=16-24-4 \\ y=-12 \end{gathered}`

Then, the parabola passes through the points (2, -12) and (4, -12). Connecting these points and the vertex, we get the next graph:

The line y = 3 is shown in green.

From the graph, we can see that the parabola (x²-6x-4, in blue) is greater than 3 (in green) for the next values of x:

x < -1 or x > 7