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Use the vertex (h, k) and a point on the graph (x, y) to find the general form of the equation of the quadratic function.

(h, k) = (−6, −1), (x, y) = (−9, 8)

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To find the general form of the equation of the quadratic function given a vertex and a point, we can proceed as follows:

The vertex form of the equation for a quadratic function can be expressed as:

\[ f(x) = a(x - h)^2 + k \]

where
- \( a \) is a coefficient that affects the width and direction of the parabola,
- \( (h, k) \) is the vertex of the parabola.

We've been given the vertex \( (h, k) = (-6, -1) \) and a point on the graph \( (x, y) = (-9, 8) \).

Let's use the vertex form and plug in the values of the vertex and the given point to solve for \( a \).

\[ 8 = a(-9 + 6)^2 - 1 \]
\[ 8 + 1 = a(-3)^2 \]
\[ 9 = a(9) \]
\[ a = 9 / 9 \]
\[ a = 1 \]

Now we know that \( a \) equals 1, and our vertex form equation becomes:

\[ f(x) = 1(x + 6)^2 - 1 \]

To write the equation in general form \( f(x) = ax^2 + bx + c \), we need to expand the equation:

\[ f(x) = 1(x + 6)^2 - 1 \]
\[ f(x) = (x^2 + 12x + 36) - 1 \] (Expand the square)
\[ f(x) = x^2 + 12x + 36 - 1 \]

Finally, combine the constants:

\[ f(x) = x^2 + 12x + 35 \]

So the general form of the equation of the quadratic function given the vertex \( (-6, -1) \) and the point \( (-9, 8) \) is:

\[ y = x^2 + 12x + 35 \]

Here, \( a = 1 \), \( b = 12 \), and \( c = 35 \).

User Robyne
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