Question

Consider the following quadratic function Part 3 of 6: Find the x-intercepts. Express it in ordered pairs.Part 4 of 6: Find the y-intercept. Express it in ordered pair.Part 5 of 6: Determine 2 points of the parabola other than the vertex and x, y intercepts.Part 6 of 6: Graph the function

Answer 1

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:

$y=ax^2+bx+c$

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:

$x_v=-\frac{b}{2a}$

In this problem, we have:

$y=-x^2-6x-13$

Then:

a = -1

b = -6

We write now:

$x_v=-\frac{-6}{2(-1)}=-\frac{-6}{-2}=-\frac{6}{2}=-3$

Part 3:

For this part, we need to find the x-intercepts. This is, when y = 0:

$-x^2-6x-13=0$

To solve this, we can use the quadratic formula:

$x_{1,2}=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot(-1)\cdot(-13)}}{2(-1)}$

And solve:

$x_{1,2}=\frac{6\pm\sqrt{36-52}}{-2}$$x_{1,2}=\frac{-6\pm\sqrt{-16}}{2}$

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:

$y=-0^2-6\cdot0-13=-13$

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:

$x=1\Rightarrow y=-1^2-6\cdot1-13=-1-6-13=-20$$x=-1\Rightarrow y=-(-1)^2-6\cdot(-1)-13=-1+6-13=-8$

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:

$y=-(-3)^2-6(-3)-13=-9+18-13=-4$

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: