1. You should notice that the vertex of the quadratic function denotes the critical point of the function, hence you may use the derivative of the function to find the critical value and then the critical point.

Differentiating the function with respect to x yields:

f'(x) = -6x - 12

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1. You should notice that the vertex of the quadratic function denotes the critical point of the function, hence you may use the derivative of the function to find the critical value and then the critical point.

Differentiating the function with respect to x yields:

f'(x) = -6x - 12

You need to find the solution to the equation f'(x)=0 => -6x - 12 = 0

-6x = 12 => x = -2

The critical value of the function is x=-2.

Substituting -2 for x in equation of function yields:

`f(-2) = -3(-2)^2 - 12*(-2) - 7=gt f(-2) = -12 + 24 - 7`

f(-2) = 5

**Hence, the vertex of the quadratic function is (-2,5)**

2. **The axis of symmetry passes through the vertex of graph of quadratic function and it is parallel to y axis.**

3. You need to remember that the graph of a function intercepts y axis at x = 0, hence you need to substitute 0 for x in equation of the function such that f(0)=-7.

**Hence, the graph of the function intercepts y axis at (0,-7).`**

``4. You need to find another point of the graph using the y intercept and the axis of symmetry, hence you need to draw a line that is parallel to x axis, it starts from y intercept and it intercepts the axis of symmetry.

**Hence, the y intercept point is mirrored and the image of this point is (-4,-7).**

5.

The red parabola represents the graph of the function and the black line is the line that passes through y intercept (0,-7) and the axis of symmetry. Notice that if you draw an orthogonal projection from the intercepting point between the black line and parabola, this projection falls at x=-4.