Chapter EP, Problem 10.6.7EP

To determine

To write: The equation in standard form, state the graph of the equation is a parabola, circle, ellipse, or hyperbola and then graph the equation.

Answer to Problem 10.6.7EP

The standard form of equation is $y=-3{\left(x-2\right)}^2-5$ , the equation is Parabola and graph is shown in figure (1).

Explanation of Solution

Given:

The equation is $3x^2+y=12x-17$ .

Calculation:

Rearrange the equation in the form of perfect square as follows.

$\begin{array}{c}3x^2+y=12x-17\\ y=-3x^2+12x-17\\ y=-3\left(x^2-4x\right)-17\\ y=-3\left(x^2-4x+4-4\right)-17\end{array}$

Simplify further.

$\begin{array}{c}y=-3{\left(x-2\right)}^2+12-17\\ y=-3{\left(x-2\right)}^2-5\end{array}$

The general form of Parabola equation is $y=a\left(x-k^2\right)+h$ .

Therefore, the given equation is Parabola.

The graph of the equation $3x^2+y=12x-17$ or $y=-3{\left(x-2\right)}^2-5$ is as follows.

Figure (1)

Therefore, the standard form of equation is $y=-3{\left(x-2\right)}^2-5$ , the equation is Parabola and graph is shown in figure (1).