McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
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Exercise 35 Page 218

Identify the vertex first. Then use it to find the axis of symmetry.

Vertex Form: y=- 12(x+2)^2+14
Vertex: (-2,14)
Axis of Symmetry: x=-2
Opening: Opens Down
Graph:

Practice makes perfect

In order to graph the function we will rewrite the function into vertex form; identify a, h, and k; plot the vertex and points; and then put it together.

Rewriting Function

We have a quadratic function written in standard form, and we want to rewrite it in vertex form. Standard Form:& y= ax^2+ bx+c Given Equation:& y= - 1/2x^2 -2x+12 In the given equation, a= - 12, b= -2, and c=12. Let's now recall the vertex form of a quadratic function. Vertex Form: y=a(x-h)^2+kIn this equation, a is the leading coefficient of the quadratic function, and the point (h,k) is the vertex of the parabola. By substituting our given values for a and b into the expression - b2a, we can find h.
- b/2a
- -2/2( - 12)
â–Ľ
Simplify
- -2/-1
- 2/1
-2
So far, we know that the vertex lies at (-2,k). To find the y-coordinate k, we will substitute -2 for x in the given function.
y=- 1/2x^2-2x+12
y=- 1/2( -2)^2-2( -2)+12
â–Ľ
Simplify right-hand side
y=- 1/2(4)-2(-2)+12
y=- 4/2-(-4)+12
y=- 2-(-4)+12
y=-2+4 +12
y=14
Therefore, the (h,k) coordinate pair of the vertex is ( -2,14). Moreover, since we already know that a= - 12, we can rewrite the given function in vertex form. y= - 12 (x-( -2))^2+14 ↔ y= - 12 (x+ 2)^2+14

Identify a, h, and k

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:& f(x)= a(x- h)^2+k Function:& f(x)= - 12(x+ 2)^2+14 We can see that a= - 12, h= -2, and k=14. Since a is less than 0, the parabola will open downwards.

Plotting the Vertex and Points

Let's now plot the vertex ( h,k) and draw the axis of symmetry x= h. Since we already know the values of h and k, we know that the vertex is ( -2,14). Therefore, the axis of symmetry is the vertical line x= -2.

We will now plot a point on the curve by choosing an x-value and calculating its corresponding y-value. Let's try x=2.
y=- 12(x+2)^2+14
y=- 12(2+2)^2+14
â–Ľ
Simplify right-hand side

\Add

y=- 12(4)^2+14
y=- 12(16)+14
y=- 162+14
y=- 8 + 14
y=6
When x=2, we have y=6. Thus, the point (2,6) lies on the curve. Let's plot this point and reflect it across the axis of symmetry.

Note that both points have the same y-coordinate.

Putting it Together

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