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y = 3(x + 3)(x + 1)Intercept form find the vertex

1 Answer

6 votes

Answer:

(-2, -3)

Step-by-step explanation:

If we have an equation with the form:

f(x) = ax² + bx + c

The vertex of the parabola is the point (-b/2a, f(-b,2a)).

So, to find the vertex, we need to write the equation y = 3(x + 3)(x + 1) as:


\begin{gathered} y=3(x+3)(x+1) \\ y=3(x^2+3x+x+3) \\ y=3(x^2+4x+3) \\ y=3x^2+12x+9 \end{gathered}

Then, the value of a is 3 and the value of b is 12. It means that -b/2a is equal to:


-(b)/(2a)=-(12)/(2(3))=-(12)/(6)=-2

And f(-b/2a) = f(-2) is equal to:


\begin{gathered} y=3x^2+12x+9 \\ f(-2)=3(-2)^2+12(-2)+9 \\ f(-2)=3(4)-24+9 \\ f(-2)=12-24+9 \\ f(-2)=-3 \end{gathered}

Therefore, the vertex is the point (-2, -3)

User Byron Singh
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