Consider the following function.q(x) = -(x + 1)^2 - 6Step 3 of 4: Find two points on the graph of the parabola other than the vertex and x-intercepts.



Answer :

Answer:

(-2, - 7), (1, -10)

Explanation:

Two find any two points, we put in our desired values of x and solve to get the output

Let us choose x = - 2. Putting this value into the equation gives

[tex]q(-2)=-(-2+1)^2-6[/tex]

simplifying the above gives

[tex]\begin{gathered} q(-2)=-1(-1)^2-6 \\ q(-2)=-1-6 \\ q(-2)=-7 \end{gathered}[/tex]

Hence, we have the point (-2, -7).

Now we choose x = 1 and put it into the equation:

[tex]q(1)=-1(1+1)^2-6[/tex]

[tex]q(1)=-4-6[/tex][tex]q(1)=-10[/tex]

Hence, we have the point ( 1, -10).

To summarise, the two points we got are:

[tex](1,-10),(-2,-7)[/tex]