State the vertex, intercepts, and domain & range for each quadratic.f(x)=−4(x+23)2−3f(x)=-4\left(x+\frac23\right)^2-3

Vertex: (−23,−3)\left(-\frac23,-3\right); xx-intercept: None; yy-intercept: (0,−439)\left(0,-\frac{43}{9}\right); Domain: (−∞,∞)\left(-\infty,\infty\right); Range: (−∞,−3]\left(-\infty,-3\right\rbrack

State the vertex, intercepts, and domain & range for each quadratic. f ( x ) = − 4 ( x + 2 3 ) 2 − 3 f(x)=-4\left(x+\frac23\right)^2-3

For this problem, we need to state the vertex, intercepts, and the domain and range for this quadratic. Negative four times x plus two thirds squared minus three. So let's start with the vertex. We know that our vertex can be found at h comma k, where here our h value is negative two thirds, because remember we need to rewrite this as x minus h, so x minus negative two thirds, and our k value is negative three. So that means our vertex is at negative two thirds, comma negative three. Next, let's take a look at our x intercepts. To find our x intercepts, we're gonna take our equation negative four times x plus two third squared minus three. We're gonna set it equal to zero and solve for x. Adding three to both sides, we get negative four times x plus two thirds squared equals three. Divide both sides by negative four, we have x plus two thirds squared equals negative three fourths. Using our square root property, we square root both sides to get x plus two thirds is equal to positive and negative square root of negative three fourths. And I can actually stop here because our value under the square root is negative, which means that this solution is going to be imaginary, which means that we have no real x intercepts because they're imaginary or complex. Next, let's take a look at our y intercept, which we know we can get by plugging in zero for x into our equation. So we'll have it negative four times zero plus two thirds squared minus three. We're gonna evaluate this to get y. Zero plus two thirds is two thirds, and two thirds squared is four ninths. So we'll have negative four times four ninths minus three equals y. Negative four times four is 16, so we have negative 16 ninths minus three equals y. Getting a common denominator, we'll rewrite three as 27 divided by nine, and negative 16 minus 27 is negative 43, over nine, is equal to y. So when x equals zero, y equals negative 43 nights. Now, let's take a look at our domain, which is always the same for parabolas, negative infinity to positive infinity, but our range depends on the y value of our vertex. Here, that y value is negative three. And since our a value is negative, that means that this parabola is opening downward, which means our vertex is a high point for this parabola. So our range is going to be from negative infinity to negative three.

State the vertex, intercepts, and domain & range for each quadratic.f(x)=−4(x+23)2−3f(x)=-4\left(x+\frac23\$$right)^2-3 $$

Vertex: (−23,−3)\left(-\frac23,-3\right); xx-intercept: None; yy-intercept: (0,−439)\left(0,-\frac{43}{9}\right); Domain: (−∞,∞)\left(-\infty,\infty\right); Range: (−∞,−3]\left(-\infty,-3\right\rbrack

Vertex: (−23,−3)\left(-\frac23,-3\right)(−32,−3); xxx-intercept: (0,0)\left(0,0\right) ; yyy-intercept: (0,−439)\left(0,-\frac{43}{9}\right)(0,−943); Domain: (−∞,∞)\left(-\infty,\infty\right)(−∞,∞); Range: (−∞,−3]\left(-\infty,-3\right\rbrack(−∞,−3]

Vertex: (0,−439)\left(0,-\frac{43}{9}\right); xxx-intercept: None; yyy-intercept: (0,−3)\left(0,-3\right); Domain: (−∞,∞)\left(-\infty,\infty\right)(−∞,∞); Range: (−∞,0]\left(-\infty,0\right\rbrack

Vertex: (0,−439)\left(0,-\frac{43}{9}\right); xxx-intercept: None; yyy-intercept: (0,−439)\left(0,-\frac{43}{9}\right); Domain: (−∞,∞)\left(-\infty,\infty\right)(−∞,∞); Range: (−∞,−3]\left(-\infty,-3\right\rbrack

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1. Review of Real Numbers2h 39m

Simplifying Fractions
30m

Evaluating Exponents
29m

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15m

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16m

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9m

The Distributive Property
17m

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40m

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41m

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30m

Solving Linear Equations
1h 14m

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1h 11m

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37m

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21m

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59m

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43m

4. Graphing Linear Equations in Two Variables3h 17m

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44m

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24m

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23m

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44m

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38m

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22m

5. Systems of Linear Equations1h 43m

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41m

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15m

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45m

6. Exponents and Polynomials3h 25m

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10m

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13m

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18m

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18m

Negative Exponents
28m

Simplifying Exponential Expressions Using All Exponent Rules
6m

Intro to Polynomials
21m

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20m

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33m

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34m

7. Factoring2h 42m

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37m

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11m

Factoring Trinomials of the Form ax² + bx + c
37m

Factoring Special Products
33m

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41m

8. Rational Expressions and Equations3h 13m

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39m

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25m

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19m

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32m

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32m

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44m

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42m

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38m

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39m

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28m

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23m

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57m

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15m

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5m

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24m

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27m

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24m

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27m

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19m

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23m

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18m

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51m

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The Square Root Property
25m

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26m

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40m

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1h 28m

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25m

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13m

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26m

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32m

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15m

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10m

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Introduction to Sequences
40m

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27m

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26m

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11m

12. Quadratic Equations and Functions

Graphing Quadratic Equations

Beginning & Intermediate Algebra

12. Quadratic Equations and Functions

Graphing Quadratic Equations

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Multiple Choice

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

Vertex: $(- \frac{2}{3}, - 3)$ ; $x$ -intercept: None; $y$ -intercept: $(0, - \frac{43}{9})$ ; Domain: $(- \infty, \infty)$ ; Range: $(- \infty, - 3]$

Vertex: $$\left$(-$\frac$23,-3$\right$)$ ; $x$ -intercept: $open paren 0 comma 0 close paren$ ; $y$ -intercept: $$\left$(0,-$\frac{43}{9}\]\right$)$ ; Domain: $$\left$(-$\infty$,$\infty\]\right$)$ ; Range: $\left\)(-$\infty$,-3\(\right\]\rbrack$

Vertex: $(0, - \frac{43}{9})$ ; $x$ -intercept: None; $y$ -intercept: $open paren 0 comma negative 3 close paren$ ; Domain: $$\left$(-$\infty$,$\infty\]\right$)$ ; Range: $open paren negative infinity comma 0 close bracket$

Vertex: $(0, - \frac{43}{9})$ ; $x$ -intercept: None; $y$ -intercept: $(0, - \frac{43}{9})$ ; Domain: $$\left$(-$\infty$,$\infty\]\right$)$ ; Range: $(- \infty, - 3]$

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Identify the vertex from the given quadratic function in vertex form $f(x) = -4\left(x + \frac{2}{3}\right)^2 - 3$. The vertex is given by the point $\left(-\frac{2}{3}, -3\right)$, where the expression inside the squared term is zero and the constant term outside is the y-coordinate.

Determine the x-intercepts by setting $f(x) = 0$ and solving for $x$: $0 = -4\left(x + \frac{2}{3}\right)^2 - 3$. Add 3 to both sides and then divide by -4, then take the square root to solve for $x$. If no real solutions exist, then there are no x-intercepts.

Find the y-intercept by evaluating $f(0)$, which means substituting $x=0$ into the function: $f(0) = -4\left(0 + \frac{2}{3}\right)^2 - 3$. Simplify this expression to get the y-intercept point $(0, f(0))$.

State the domain of the quadratic function. Since it is a polynomial function, the domain is all real numbers, expressed as $\left(-\infty, \infty\right)$.

Determine the range based on the vertex and the leading coefficient. Since the coefficient of the squared term is negative ($-4$), the parabola opens downward, so the range is all $y$ values less than or equal to the y-coordinate of the vertex. Express the range as $\left(-\infty, -3\right]$.

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State the vertex, intercepts, and domain & range for each quadratic.f(x)=−4(x+23)2−3f(x)=-4\(\left\)(x+\(\frac\)23\(\right\))^2-3

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