Graph f(x)=x^2-4x-5

Problem

ƒ(x)=x2−4*x−5

Solution

1. Identify the type of function and its orientation. This is a quadratic function in the form ƒ(x)=a*x2+b*x+c where a=1 b=−4 and c=−5 Since a>0 the parabola opens upward.

2. Find the vertex using the formula x=(−b)/(2*a)

x=(−(−4))/(2*(1))

x=2

Substitute x=2 into the function to find the ycoordinate:

ƒ(2)=(2)2−4*(2)−5

ƒ(2)=4−8−5

ƒ(2)=−9

The vertex is (2,−9)

3. Find the y-intercept by evaluating ƒ(0)

ƒ(0)=0−4*(0)−5

ƒ(0)=−5

The yintercept is (0,−5)

4. Find the x-intercepts by setting ƒ(x)=0 and factoring.

x2−4*x−5=0

(x−5)*(x+1)=0

x=5

x=−1

The xintercepts are (5,0) and (−1,0)

5. Plot the points and draw a smooth curve. Use the vertex (2,−9) the yintercept (0,−5) and the xintercepts (−1,0) and (5,0) to sketch the parabola.

Final Answer

ƒ(x)=(x−2)2−9

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