Question

sing the Quadratic Formula, find the solution to 3x2+18x+24=0

Responses

x=−2

x=4, x=2

x=−2, x=−4

x=−4

Answer 1

Which of the following is a solution to the quadratic equation −x2+x=−10
? Assume that the solution has been rounded to the nearest hundredth, if applicable.
Responses

x=0.27

x=1.7

x=3.7

x=0.77

Answer 2

Which of the following coordinate pairs is a point on the parabola y=−4x^2−53x−56

Responses

(−4,−56)

(−1,−7)

(−1, 13)

(−4, 220)

Answer 3

Determine the zero(s) of the quadratic function y=−5x2+16x

Responses

(0, 0) (3.2, 0)

There are no zeros.


(−3.2, 0)

(0, 5) (0, 16)

Answer 4

identify the vertex of y=4(x−25)^2−61

Responses

(−100, −61)

(−25, −61)

(25, -61)

(4, −25)

Answer 5

What are the key points on the graph of y=x^2−2x−120

? Name the vertex, x-intercepts, and y-intercepts.

x-intercepts: (, ) and (1

Answer 6

To solve the quadratic equation \(3x^2 + 18x + 24 = 0\) using the Quadratic Formula, we first identify the coefficients \(a\), \(b\), and \(c\) in the standard form \(ax^2 + bx + c = 0\):

- \(a = 3\)
- \(b = 18\)
- \(c = 24\)

The Quadratic Formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Now, we can substitute the values of \(a\), \(b\), and \(c\) into the formula:

1. Calculate the discriminant (\(b^2 - 4ac\)):
\[
b^2 - 4ac = 18^2 - 4 \cdot 3 \cdot 24
\]
\[
= 324 - 288
\]
\[
= 36
\]

2. Substitute the values into the quadratic formula:
\[
x = \frac{-18 \pm \sqrt{36}}{2 \cdot 3}
\]
\[
= \frac{-18 \pm 6}{6}
\]

3. Calculate the two possible solutions:
- First solution:
\[
x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2
\]
- Second solution:
\[
x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4
\]

Thus, the solutions to the equation \(3x^2 + 18x + 24 = 0\) are:

\[
x = -2 \quad \text{and} \quad x = -4
\]

The correct response is:

**x=−2, x=−4**.

Answer 7

To find the solution to the quadratic equation \(-x^2 + x = -10\), let's first rearrange it into standard form:

\[
-x^2 + x + 10 = 0
\]

To simplify the calculations, we can multiply the entire equation by -1:

\[
x^2 - x - 10 = 0
\]

Now, we'll use the Quadratic Formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our equation, we have:

- \(a = 1\)
- \(b = -1\)
- \(c = -10\)

Now, we can calculate the discriminant (\(b^2 - 4ac\)):

\[
b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot (-10) = 1 + 40 = 41
\]

Now substitute into the Quadratic Formula:

\[
x = \frac{-(-1) \pm \sqrt{41}}{2 \cdot 1}
\]
\[
x = \frac{1 \pm \sqrt{41}}{2}
\]

Calculating the two possible solutions:

1. First solution:
\[
x = \frac{1 + \sqrt{41}}{2}
\]

2. Second solution:
\[
x = \frac{1 - \sqrt{41}}{2}
\]

Now we can approximate \(\sqrt{41}\):

\[
\sqrt{41} \approx 6.403
\]

Now, calculating the first solution:

\[
x \approx \frac{1 + 6.403}{2} = \frac{7.403}{2} \approx 3.7015 \approx 3.70
\]

Now calculating the second solution:

\[
x \approx \frac{1 - 6.403}{2} = \frac{-5.403}{2} \approx -2.7015
\]

Since we are looking for the solution rounded to the nearest hundredth, the solution \(x \approx 3.70\) is relevant here.

Thus, the correct response is:

**x=3.7**.

Answer 8

To determine which coordinate pairs are points on the parabola defined by the equation \(y = -4x^2 - 53x - 56\), we need to substitute the x-coordinates of each of the given pairs into the equation and check if the resulting y-value matches the y-coordinate of the pair.

The given pairs are:

1. \((-4, -56)\)
2. \((-1, -7)\)
3. \((-1, 13)\)
4. \((-4, 220)\)

Let's calculate \(y\) for \(x = -4\) and \(x = -1\):

### 1. For \((−4,−56)\):
Substituting \(x = -4\) into the equation:
\[
y = -4(-4)^2 - 53(-4) - 56
\]
\[
= -4(16) + 212 - 56
\]
\[
= -64 + 212 - 56
\]
\[
= 148 \quad (\text{not } -56)
\]

### 2. For \((−1,−7)\):
Substituting \(x = -1\):
\[
y = -4(-1)^2 - 53(-1) - 56
\]
\[
= -4(1) + 53 - 56
\]
\[
= -4 + 53 - 56
\]
\[
= -7 \quad (\text{matches } -7)
\]

### 3. For \((−1, 13)\):
Substituting \(x = -1\):
\[
y = -4(-1)^2 - 53(-1) - 56
\]
Using the same calculation as above, we have:
\[
y = -7 \quad (\text{not } 13)
\]

### 4. For \((−4, 220)\):
We already calculated \(y\) for \(x = -4\) and found \(y = 148\) (not \(220\)).

### Summary of Results
1. \((-4, -56)\) does not work.
2. \((-1, -7)\) **is a point on the parabola**.
3. \((-1, 13)\) does not work.
4. \((-4, 220)\) does not work.

Thus, the correct coordinate pair that is a point on the parabola is:

**(−1,−7)**.

Answer 9

To find the zeros of the quadratic function \(y = -5x^2 + 16x\), we need to set the equation to zero and solve for \(x\):

\[
-5x^2 + 16x = 0
\]

We can factor out \(x\):

\[
x(-5x + 16) = 0
\]

Setting each factor equal to zero gives us:

1. \(x = 0\)
2. \(-5x + 16 = 0\)

Solving the second equation for \(x\):

\[
-5x + 16 = 0 \quad \Rightarrow \quad -5x = -16 \quad \Rightarrow \quad x = \frac{16}{5} = 3.2
\]

Thus, the zeros of the function are:

\[
x = 0 \quad \text{and} \quad x = 3.2
\]

The coordinates of the zeros in terms of (x, y) are:

\[
(0, 0) \quad \text{and} \quad (3.2, 0)
\]

So the correct response from the options provided is:

**(0, 0) (3.2, 0)**.

Answer 10

The vertex form of a quadratic function is given by the equation:

\[
y = a(x - h)^2 + k
\]

where \((h, k)\) is the vertex of the parabola.

In your function \(y = 4(x - 25)^2 - 61\), we can identify the values:
- \(h = 25\)
- \(k = -61\)

Thus, the vertex of the function is:

\[
(25, -61)
\]

Therefore, the correct response is:

**(25, -61)**.

Answer 11

To find the key points on the graph of the quadratic function \(y = x^2 - 2x - 120\), we will determine the vertex, x-intercepts, and y-intercept.

### 1. Finding the Vertex

The vertex of a parabola given in standard form \(y = ax^2 + bx + c\) can be found using the vertex formula:

\[
h = -\frac{b}{2a} \quad \text{and} \quad k = f(h)
\]

In our equation, the coefficients are:
- \(a = 1\)
- \(b = -2\)
- \(c = -120\)

Calculating \(h\):

\[
h = -\frac{-2}{2 \cdot 1} = \frac{2}{2} = 1
\]

Now, substitute \(h\) back into the equation to find \(k\):

\[
k = f(1) = 1^2 - 2(1) - 120 = 1 - 2 - 120 = -121
\]

Thus, the vertex is:

\[
(1, -121)
\]

### 2. Finding the X-Intercepts

To find the x-intercepts, we set \(y = 0\):

\[
0 = x^2 - 2x - 120
\]

Now we can factor the quadratic equation:

\[
0 = (x - 12)(x + 10)
\]

Setting each factor to zero gives us the x-intercepts:

1. \(x - 12 = 0 \Rightarrow x = 12\)
2. \(x + 10 = 0 \Rightarrow x = -10\)

Thus, the x-intercepts are:

\[
(12, 0) \quad \text{and} \quad (-10, 0)
\]

### 3. Finding the Y-Intercept

To find the y-intercept, we set \(x = 0\):

\[
y = 0^2 - 2(0) - 120 = -120
\]

This gives the y-intercept:

\[
(0, -120)
\]

### Summary of Key Points
- **Vertex**: \((1, -121)\)
- **X-Intercepts**: \((12, 0)\) and \((-10, 0)\)
- **Y-Intercept**: \((0, -120)\)

So putting it all together, the key points on the graph of \(y = x^2 - 2x - 120\) are:

- Vertex: **(1, -121)**
- X-Intercepts: **(12, 0)** and **(-10, 0)**
- Y-Intercept: **(0, -120)**