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)(−32,−3); xxx-intercept: None; 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]

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)(−32,−3); xxx-intercept: None; 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: (−23,−3)\left(-\frac23,-3\right)(−32,−3); xxx-intercept: (0,0)\left(0,0\right)(0,0) ; 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)(0,−943); xxx-intercept: None; yyy-intercept: (0,−3)\left(0,-3\right)(0,−3); Domain: (−∞,∞)\left(-\infty,\infty\right)(−∞,∞); Range: (−∞,0]\left(-\infty,0\right\rbrack(−∞,0]

Vertex: (0,−439)\left(0,-\frac{43}{9}\right)(0,−943); xxx-intercept: None; 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]

10. Quadratic Equations and Functions

Graphing Quadratic Equations

Intermediate Algebra

10. 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: $$\left$(-$\frac$23,-3$\right$)$ ; $x$ -intercept: None; $y$ -intercept: $$\left$(0,-$\frac{43}{9}\]\right$)$ ; Domain: $$\left$(-$\infty$,$\infty\]\right$)$ ; Range: $\left\)(-$\infty$,-3\(\right\]\rbrack$

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

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

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

Verified step by step guidance

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 $\left(h, k\right)$ where the function is written as $f(x) = a(x - h)^2 + k$. Here, rewrite the expression inside the parentheses to match the form $(x - h)$ and determine $h$ and $k$.

Determine the $x$-intercepts by setting $f(x) = 0$ and solving the equation $0 = -4\left(x + \frac{2}{3}\right)^2 - 3$. Since the quadratic is negative and shifted down, check if the equation has real solutions by isolating the squared term and analyzing the resulting expression.

Find the $y$-intercept by evaluating $f(0)$. Substitute $x = 0$ into the function and simplify to get the $y$-intercept point $(0, f(0))$.

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

Determine the range of the function. Because the parabola opens downward (coefficient of the squared term is negative), the range includes all $y$-values less than or equal to the $y$-coordinate of the vertex. Express the range accordingly using interval notation.

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Struggling with Intermediate Algebra?

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.
$f (x) = - 4 {(x + \frac{2}{3})}^{2} - 3$