McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
1. Graphing Quadratic Functions
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Exercise 55 Page 225

Identify the coordinates of the vertex.

y=x^2-4x-5

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 ( 2,-9). This means that we have h= 2 and k=-9. We can use these values to partially write our function. y= a(x-( 2))^2+(-9) ⇕ y= a(x-2)^2-9 We can see in the graph that the parabola opens upwards. Therefore, 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 (0,- 5). Since this point is on the curve, it satisfies its equation. Hence, to find the value of a, we can substitute 0 for x and - 5 for y and simplify.
y=a(x-2)^2-9
- 5=a( 0-2)^2-9
â–Ľ
Solve for a
- 5=a(- 2)^2-9
- 5=a(4)-9
4=a(4)
1=a
a=1
We found that a= 1. Now we can complete the equation of the curve. y= 1(x-2)^2-9 To obtain the standard form of the quadratic function, we simplify the right-hand side of the equation.
y=(x-2)^2-9
y=x^2-2x(2)+2^2-9
y=x^2-4x+4-9
y=x^2-4x-5