Exercise 3 Page 224

We want to draw the graph of the given quadratic function. To do so, we will rewrite it in vertex form y=a(x-h)^2+k, where a, h, and k are either positive or negative numbers. y=- (x+2)^2-7 ⇕ y=- 1(x-(- 2))^2+(- 7) To draw the graph, we will follow four steps.

1. Identify the constants a, h, and k.
2. Plot the vertex (h,k) and draw the axis of symmetry x=h.
3. Plot any point on the curve and its reflection across the axis of symmetry.
4. Sketch the curve.

   Let's get started.

   Step 1

   We will first identify the constants a, h, and k. Recall that if a<0, the parabola will open downwards. Conversely, if a>0, the parabola will open upwards. Vertex Form:& y= a(x- h)^2+ k Function:& y= - 1(x-( - 2))^2+( - 7) We can see that a= - 1, h= - 2, and k= - 7. Since a is less than 0, the parabola will open downwards.

   Step 2

   Recall that the axis of symmetry is a vertical line that passes through the vertex ( h, k). As a result, it takes form x= h. Since we already know the values of h and k, we know that the vertex is ( - 2, - 7) and the axis of symmetry is x= - 2.

   

   Step 3

   We will now plot a point on the curve by choosing an arbitrary x-value and calculating its corresponding y-value. Let's try x=0.
   y=- (x+2)^2 -7

   Substitute

   x= 0

   y=- ( 0+2)^2 -7

   ▼

   Simplify right-hand side

   IdPropAdd

   Identity Property of Addition

   y=- (2)^2 -7

   CalcPow

   Calculate power

   y=- (4)-7

   RemovePar

   Remove parentheses

   y=- 4-7

   SubTerm

   Subtract term

   y=- 11
   When x=0, we have y=- 11. Thus, the point (0,- 11) lies on the curve. Let's plot this point and its reflection across the axis of symmetry.
   

   Note that both points have the same y-coordinate.

   Step 4

   Finally, we will sketch the parabola which passes through the three points. Remember not to use a straight edge for this!