Graph f(x)=3x^2-2x-5

Problem

ƒ(x)=3*x2−2*x−5

Solution

1. Identify the function type and the coefficients. This is a quadratic function in the form ƒ(x)=a*x2+b*x+c where a=3 b=−2 and c=−5 Since a>0 the parabola opens upward.

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

x=(−(−2))/(2*(3))

x=1/3

Substitute x=1/3 into the function to find the ycoordinate.

ƒ(1/3)=3*(1/3)2−2*(1/3)−5

ƒ(1/3)=1/3−2/3−5

ƒ(1/3)=−16/3

The vertex is (1/3,−16/3)

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

ƒ(0)=3*(0)2−2*(0)−5

ƒ(0)=−5

The yintercept is (0,−5)

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

3*x2−2*x−5=0

(3*x−5)*(x+1)=0

Setting each factor to zero gives x=5/3 and x=−1
The xintercepts are (5/3,0) and (−1,0)

5. Plot the points and draw the smooth U-shaped curve (parabola) passing through the vertex (1/3,−5.33) the yintercept (0,−5) and the xintercepts (−1,0) and (1.67,0)

Final Answer

ƒ(x)=3*x2−2*x−5

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