State the vertex, axis of symmetry, and domain & range for each quadratic. f(x)=(x+14)2+3f\left(x\right)=\left(x+\frac14\right)^2+3

Vertex: (−14,3)\left(-\frac14,3\right); Axis of symmetry: x=−14x=-\frac14; Domain: (−∞,∞)\left(-\infty,\infty\right); Range: [3,∞)\left\lbrack3,\infty\right)

State the vertex, axis of symmetry, and domain & range for each quadratic. f ( x ) = ( x + 1 4 ) 2 + 3 f\left(x\right)=\left(x+\frac14\right)^2+3

this problem, we need to find the vertex, axis of symmetry, and domain and range for this quadratic. X plus one fourth, all squared, plus three. So let's start with our vertex. Here, we have two shifts. We have a horizontal shift and a vertical shift. We're gonna rewrite this to look more like our form of x minus h. So we'll rewrite it as x minus negative one fourth, which makes it easy to see that we have a shift to the left of one fourth. We also have a shift upwards of three, which means our new vertex is going to be at negative one fourth, positive three. Next, for our axis of symmetry, it always runs through the vertex and shares its value with the x value. So it's going to be the vertical line, x equals negative one fourth. Lastly, for our domain and range, we know that the domain is always the same from negative infinity to positive infinity, but our range depends on the vertex. Our vertex y value is at three, And since this quadratic is positive in front, that means that it's opening upward, so that our vertex is then a low point for our graph. So that means the range of this parabola is going to be from three to positive infinity.

State the vertex, axis of symmetry, and domain & range for each quadratic. f(x)=(x+14)2+3f\left(x\right)=\left(x+\frac14\$$right)^2+3 $$

Vertex: (−14,3)\left(-\frac14,3\right); Axis of symmetry: x=−14x=-\frac14; Domain: (−∞,∞)\left(-\infty,\infty\right); Range: [3,∞)\left\lbrack3,\infty\right)

Vertex: (−14,−3)\left(-\frac14,-3\right); Axis of symmetry: x=−14x=-\frac14x=−41; Domain: (−∞,∞)\left(-\infty,\infty\right)(−∞,∞); Range: (−∞,−3]\left(-\infty,-3\right\rbrack

Vertex: (14,−3)\left(\frac14,-3\right); Axis of symmetry: x=14x=\frac14; Domain: (−∞,∞)\left(-\infty,\infty\right)(−∞,∞); Range: [−3,∞)\left\lbrack-3,\infty\right)

Vertex: (14,3)\left(\frac14,3\right); Axis of symmetry: x=14x=\frac14; Domain: (−∞,∞)\left(-\infty,\infty\right)(−∞,∞); Range: (−∞,3]\left(-\infty,3\right\rbrack

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Graphing Quadratic Equations

Beginning & Intermediate Algebra

12. Quadratic Equations and Functions

Graphing Quadratic Equations

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Multiple Choice

State the vertex, axis of symmetry, and domain & range for each quadratic.
$f (x) = {(x + \frac{1}{4})}^{2} + 3$

Vertex: $(- \frac{1}{4}, - 3)$ ; Axis of symmetry: $x=-$\frac$14$ ; Domain: $$\left$(-$\infty$,$\infty\]\right$)$ ; Range: $(- \infty, - 3]$

Vertex: $open paren one fourth comma negative 3 close paren$ ; Axis of symmetry: $x = \frac{1}{4}$ ; Domain: $$\left$(-$\infty$,$\infty\]\right$)$ ; Range: $open bracket negative 3 comma infinity close paren$

Vertex: $open paren one fourth comma 3 close paren$ ; Axis of symmetry: $x = \frac{1}{4}$ ; Domain: $$\left$(-$\infty$,$\infty\]\right$)$ ; Range: $open paren negative infinity comma 3 close bracket$

Vertex: $open paren negative one fourth comma 3 close paren$ ; Axis of symmetry: $x = - \frac{1}{4}$ ; Domain: $(- \infty, \infty)$ ; Range: $open bracket 3 comma infinity close paren$

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Identify the quadratic function given in vertex form: $f\left(x\right)=\left(x+\frac{1}{4}\right)^2+3$.

Recall that the vertex form of a quadratic is $f\left(x\right) = a(x-h)^2 + k$, where the vertex is at $(h, k)$. Here, rewrite $x + \frac{1}{4}$ as $x - \left(-\frac{1}{4}\right)$ to identify $h = -\frac{1}{4}$ and $k = 3$.

State the vertex as $\left(-\frac{1}{4}, 3\right)$ based on the values of $h$ and $k$.

Determine the axis of symmetry, which is the vertical line passing through the vertex, given by the equation $x = h$, so here $x = -\frac{1}{4}$.

Identify the domain and range: The domain of any quadratic function is all real numbers, so $\left(-\infty, \infty\right)$. Since the parabola opens upwards (coefficient of squared term is positive), the range starts at the vertex's $y$-value and goes to infinity, so the range is $\left[3, \infty\right)$.

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State the vertex, axis of symmetry, and domain & range for each quadratic. f(x)=(x+14)2+3f\(\left\)(x\(\right\))=\(\left\)(x+\(\frac\)14\(\right\))^2+3

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State the vertex, axis of symmetry, and domain & range for each quadratic.
$f (x) = {(x + \frac{1}{4})}^{2} + 3$