McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Quadratic Inequalities
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Exercise 69 Page 213

Identify the coordinates of the vertex.

y=0.25(x+4)^2+3

Practice makes perfect
We want to write the equation of the given parabola. To do so, let's recall the vertex form of a quadratic function. y=a(x-h)^2+k In this expression, a, h, and k are either positive or negative constants. Let's start by identifying the vertex.

The vertex of this parabola has coordinates (- 4,3). This means that we have h=- 4 and k=3. We can use these values to partially write our function. y=a(x-(- 4))^2+3 ⇕ y=a(x+4)^2+3 We can see in the graph that the parabola opens upwards. Thus, a will be a positive number. To find its value, we will use the given point that is not the vertex.

We can see above that the point has coordinates (- 8,7). Since this point is on the curve, it satisfies its equation. Hence, to find the value of a, we can substitute - 8 for x and 7 for y and simplify.
y=a(x+4)^2+3
7=a( - 8+4)^2+3
â–Ľ
Solve for a
7=a(- 4)^2+3
7=a(16)+3
4=a(16)
4/16=a
1/4=a
a=1/4
a=0.25
We found that a=0.25. Now we can complete the equation of the curve. y=0.25(x+4)^2+3