What is the minimum value of $y$ if $y=3x^2+6x+9?$
The standard dorm of a auadratic equation:
y = a x² + b x + c
The vertex of a quadratic equation is either a maximum or a minimum of the function.
If leading coefficient a > 0, then the parabola opens up and vertex ts minimum.
If leading coefficient a < 0, then the parabola opens down and vertex ts maximum.
The value x = − b / 2a tells you the x value of the vertex.
In this case y = 3 x² + 6 x + 9
a = 3 , b = 6 , c = 9
a > 0, vertex is minimum
x = − b / 2a = - 6 / 2 ∙ 3 = - 6 / 6 = - 1
ymin = y(-1) = 3 ∙ ( - 1 )² + 6 ∙ ( - 1 ) + 9 = 3 ∙ 1 - 6 + 9 = 3 - 6 + 9 = 6
Y = 3x^2 + 6x + 9.
Since "A" is positive, the parabola opens upward and vertex is min. point
on the curve.
h = Xv = -B/2A = -6/(2*3) = -1.
K = Yv = Ymin. = 3x^2 + 6x + 9 = 3*(-1)^2 + 6*(-1) + 9 = 6.
V(h, k) = V(-1, 6).
Hmm, let me calculate that for you. But before I do, brace yourself, because the answer might be very small! Just kidding, it's actually quite simple. To find the minimum value of $y$, we need to locate the vertex of the parabola defined by the equation $y=3x^2+6x+9$. And guess what? The vertex occurs at the minimum point of the parabola. So, let's find the x-coordinate of the vertex first. Using the formula $x=-\frac{b}{2a}$ for a quadratic equation in the form $ax^2+bx+c$, we have $x=-\frac{6}{2\cdot3}$. Simplifying this gives us $x=-1$. Now, plugging this x-coordinate back into the equation, we get $y=3(-1)^2+6(-1)+9$. After some calculation, the minimum value of $y$ is 6! Ta-da! But hey, don't feel too bad for those small, negative values of $y$. They have their own charm!
To find the minimum value of $y$, we need to determine the vertex of the parabola represented by the equation $y=3x^2+6x+9$.
The vertex of a parabola in the form $y=ax^2+bx+c$ is given by the coordinates $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$.
In this case, $a=3$ and $b=6$, so the x-coordinate of the vertex is $-\frac{6}{2(3)} = -1$.
To find the y-coordinate, we substitute $x=-1$ into the equation $y=3x^2+6x+9$:
$y = 3(-1)^2 + 6(-1) + 9$
$y = 3 - 6 + 9$
$y = 6$
Therefore, the minimum value of $y$ is 6.
To find the minimum value of $y$, we can start by recognizing that the given expression is a quadratic function in the form of $y = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants.
In general, the minimum (or maximum) value of a quadratic function can be found using the concept of vertex. The vertex represents either the highest point (in the case of a downward-facing parabola) or the lowest point (in the case of an upward-facing parabola) on the graph of the function.
For a quadratic function in the form of $y = ax^2 + bx + c$, the $x$-coordinate of the vertex can be found using the formula: $$x = -\frac{b}{2a}$$
Plugging in the values from the given function, we get: $$x = -\frac{6}{2(3)} = -1$$
Now, to find the minimum value of $y$, we substitute this value of $x$ back into the original equation: $$y = 3(-1)^2 + 6(-1) + 9 = 3 - 6 + 9 = 6$$
Therefore, the minimum value of $y$ is 6.