Answer:
The line of symmetry is x = -3
Explanation:
Given a quadratic function, we know that the graph is a parabola. The general form of a parabola is:
[tex]y=ax^2+bx+c[/tex]
The line of symmetry coincides with the x-axis of the vertex. To find the x-coordinate of the vertex, we can use the formula:
[tex]x_v=-\frac{b}{2a}[/tex]
In this problem, we have:
[tex]y=-x^2-6x-13[/tex]
Then:
a = -1
b = -6
We write now:
[tex]x_v=-\frac{-6}{2(-1)}=-\frac{-6}{-2}=-\frac{6}{2}=-3[/tex]
Part 3:For this part, we need to find the x-intercepts. This is, when y = 0:
[tex]-x^2-6x-13=0[/tex]
To solve this, we can use the quadratic formula:
[tex]x_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot(-1)\cdot(-13)}}{2(-1)}[/tex]
And solve:
[tex]x_{1,2}=\frac{6\pm\sqrt{36-52}}{-2}[/tex][tex]x_{1,2}=\frac{-6\pm\sqrt{-16}}{2}[/tex]
Since there is no solution to the square root of a negative number, the function does not have any x-intercept. The correct option is ZERO x-intercepts.
Part 4:To find the y intercept, we need to find the value of y when x = 0:
[tex]y=-0^2-6\cdot0-13=-13[/tex]
The y-intercept is at (0, -13)
Part 5:
Now we need to find two points in the parabola. Let-s evaluate the function when x = 1 and x = -1:
[tex]x=1\Rightarrow y=-1^2-6\cdot1-13=-1-6-13=-20[/tex][tex]x=-1\Rightarrow y=-(-1)^2-6\cdot(-1)-13=-1+6-13=-8[/tex]
The two points are:
(1, -20)
(-1, -8)
Part 6:
Now, we can use 3 points to find the graph of the parabola.
We can locate (1, -20) and (-1, -8)
The third could be the vertex. We need to find the y-coordinate of the vertex. We know that the x-coordinate of the vertex is x = -3
Then, y-coordinate of the vertex is:
[tex]y=-(-3)^2-6(-3)-13=-9+18-13=-4[/tex]
The third point we can use is (-3, -4)
Now we can locate them in the cartesian plane:
And that's enough to get the full graph:
