Question

How do you graph $y = - 2{x^2} - 4x + 6$?

Answer 1

Hint: According to the question we have to determine the graph for $y = - 2{x^2} - 4x + 6$. So, first of all we have to determine the value of $a,b$and $c$ by comparing the given equation as $y = - 2{x^2} - 4x + 6$ to the standard form of the equation that is $y = a{x^2} + bx + c$.
Now, we have to check that the coefficient of ${x^2}$ is positive or negative. If the coefficient of ${x^2}$ is positive then the parabola should open up, if the coefficient of ${x^2}$ is negative then the parabola should open down.
Now, we have to find the value of $x$ with the help of the formula given below.

Formula used:
$ \Rightarrow x = \dfrac{{ - b}}{{2a}}............................(A)$
Where, b is the coefficient of $y$ in the standard form of the equation that is $y = a{x^2} + bx + c$ and c is the constant term in the standard form of the equation that is $y = a{x^2} + bx + c$.
Now, we have to put the value of $x$ obtained from the formula (A) in the given equation $y = - 2{x^2} - 4x + 6$ to get the value of $y$. So, we get the vertex of the parabola graph that is $\left( {x,y} \right)$.
Now, we have to assume two values of $x$ that are 1,-2 to get the two values of $y$ respectively.
Now, we have to check the symmetry of parabola and plot the graph of the given parabola $y = - 2{x^2} - 4x + 6$ on the graph paper.

Complete step-by-step answer:
Step 1: First of all we have to compare the given equation as $y = - 2{x^2} - 4x + 6$ to the standard form of the equation that is $y = a{x^2} + bx + c$.
$ \Rightarrow a = - 2,b = - 4$, $c = 6$
Now, we can see that the value of $a$ is negative so the parabola should open down.
Step 2: Now, we have to find the value of $x$ with the help of the formula (A) as mentioned in the solution hint.
$
   \Rightarrow x = \dfrac{{ - \left( { - 4} \right)}}{{2\left( { - 2} \right)}} \\
   \Rightarrow x = \dfrac{4}{{ - 4}} \\
   \Rightarrow x = - 1 \\
 $
Step 3: Now, we have to put the value of $x$ obtained from the solution step 2 in the given equation of parabola that is$y = - 2{x^2} - 4x + 6$.
$
   \Rightarrow y = - 2{\left( { - 1} \right)^2} - 4\left( { - 1} \right) + 6 \\
   \Rightarrow y = - 2 + 4 + 6 \\
   \Rightarrow y = 8 \\
 $
Now, we have to get the vertex of the parabola that is $\left( {x,y} \right) = \left( { - 1,8} \right)$
Step 4: Now, we have to assume two values of $x$ that is 1,-2 to get the two values of $y$ respectively.
So, at $x = 1$the value of $y$ from the given equation as $y = - 2{x^2} - 4x + 6$.
$
   \Rightarrow y = - 2{\left( 1 \right)^2} - 4\left( 1 \right) + 6 \\
   \Rightarrow y = - 2 - 4 + 6 \\
   \Rightarrow y = 0 \\
 $
So, at $x = - 2$the value of $y$from the given equation as$y = - 2{x^2} - 4x + 6$.
$
   \Rightarrow y = - 2{\left( { - 2} \right)^2} - 4\left( { - 2} \right) + 6 \\
   \Rightarrow y = - 8 + 8 + 6 \\
   \Rightarrow y = 6 \\
 $
Step 5: Now, we have to get two points on the parabola that is $\left( {1,0} \right)$ and $\left( { - 2,6} \right)$,a vertex as $\left( { - 1,8} \right)$
Now, we have to take the symmetry of parabola as mirror image points of $\left( {1,0} \right)$ and $\left( { - 2,6} \right)$ are $\left( { - 2,0} \right)$and $\left( {1,6} \right)$respectively.
Step 6: Now, we have to plot the graph on the graph paper.

Final solution: Hence, with the help of the vertex and points of intersection of the parabola we have determined the graph for the given parabola which is as mentioned below:

Note:
It is necessary that we have to check that the coefficient of ${x^2}$ is positive or negative. If the coefficient of ${x^2}$ is positive then the parabola should open up, if the coefficient of ${x^2}$ is negative then the parabola should open down.
We have to assume two values of $x$ that is 1,-2 to get the two values of $y$ respectively.