Exercise 37 Page 100

Consider the general expression of a quadratic function, y=ax^2+ bx+ c, where a ≠ 0. Let's note three things we can learn from this equation.

• The y-intercept is c.
• The equation of the axis of symmetry is x=- b2a.
• The x-coordinate of the vertex is - b2a.

We will start by identifying the values of a, b, and c. y=- x^2-6x-5 ⇕ y=(-1)x^2+( - 6)x+( - 5)We can see that a = - 1, b = - 6, and c = - 5. Since the y-intercept is given by the value of c, we know that the y-intercept is - 5. Let's now substitute a=- 1 and b=- 6 into - b2a to find the axis of symmetry.

The equation of the axis of symmetry is x=- 3.
To find the vertex of the parabola, we will need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: ( - b/2a, f(- b/2a ) ) When determining the axis of symmetry, we found that - b2a=- 3. Therefore, the x-coordinate of the vertex is - 3 and the y-coordinate is f(- 3). To find this value, substitute our x-coordinate for x in the given equation.

The vertex of the parabola is (- 3,4).