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Which are the intercept points and vertex point for the function ƒ(x) = x^2 + 5x + 6?

User Uhs
by
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1 Answer

5 votes

Answer:

The intercept points are:


\begin{gathered} (-3,0) \\ (-2,0) \end{gathered}

The vertex is:


(-(5)/(2),-(1)/(4))

Explanation:

To find the intercept points, we equal the function to zero and solve for x, as following:


\begin{gathered} x^2+5x+6=0 \\ \rightarrow(x+3)(x+2)=0 \\ \\ \rightarrow x+3=0\Rightarrow x_1=-3 \\ \\ \rightarrow x+2=0\Rightarrow x_2=-2 \end{gathered}

Now, we know that the intercept points have an x-value of -3 and -2. Since we're talking about intercepts, the y-values will be 0.

Therefore, we can conlcude that the intercept points are:


\begin{gathered} (-3,0) \\ (-2,0) \end{gathered}

The vertex of any given quadratic function in the form:


f(x)=ax^2+bx+c

is:


(-(b)/(2a),f(-(b)/(2a)))

This way, for the given function, we'll have that the vertex point is:


(-(5)/(2),-(1)/(4))

User Miquelle
by
8.7k points

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