Find the vertex of the graph of the quadratic function y = f(x) = 4 + 12x - 3x^2.
Understand the Problem
The question asks us to find the vertex of the graph of the given quadratic function. To solve this, we can rewrite the quadratic function in vertex form, which is (y = a(x - h)^2 + k), where ((h, k)) is the vertex of the parabola. Alternatively, we can use the formula (h = -b / 2a) to find the x-coordinate of the vertex, and then substitute this value back into the original equation to find the y-coordinate.
Answer
$(2, 16)$
Answer for screen readers
$(2, 16)$
Steps to Solve
- Rewrite the quadratic function in standard form
The given quadratic function is $y = 4 + 12x - 3x^2$. We rewrite it in the standard form $y = ax^2 + bx + c$, which gives us $y = -3x^2 + 12x + 4$.
- Find the x-coordinate of the vertex
The x-coordinate of the vertex, $h$, is given by the formula $h = -\frac{b}{2a}$. In our case, $a = -3$ and $b = 12$. Thus, $$h = -\frac{12}{2(-3)} = -\frac{12}{-6} = 2$$
- Find the y-coordinate of the vertex
To find the y-coordinate of the vertex, $k$, substitute $x = h = 2$ into the equation: $$k = -3(2)^2 + 12(2) + 4 = -3(4) + 24 + 4 = -12 + 24 + 4 = 16$$
- State the vertex
The vertex of the graph is $(h, k) = (2, 16)$.
$(2, 16)$
More Information
The graph of the quadratic function $y = -3x^2 + 12x + 4$ is a parabola that opens downward since the leading coefficient $a = -3$ is negative. The vertex $(2, 16)$ is the highest point on this parabola.
Tips
A common mistake is to incorrectly identify the coefficients $a$, $b$, and $c$ when the quadratic is not in standard form. Make sure to rewrite the quadratic in the form $ax^2 + bx + c$ before identifying these values. Another common mistake is an arithmetic error when calculating the x and y coordinates of the vertex.
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