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Question

Use the image to answer the question.
Four points are plotted on a coordinate plane, forming a parabolic curve.

The quadratic function f(x)=x2+2x−15
is shown in the graph. Determine the domain and range of this function.

(1 point)
Responses

domain R
; range y≥16
domain the real numbers ; range y is greater than or equal to 16

domain R
; range y≤−16
domain negative infinity less-than-or-equal-to x less-than-or-equal-to infinity ; range y is less than or equal to negative 16

domain R
; range y≥−16
domain the real numbers ; range y is greater than or equal to negative 16

domain R
; range R

Use the image to answer the question.

A coordinate plane shows the x-axis and the y-axis ranging from negative 5 to 5, in increments of 0.5 units, but labeled at every unit. An upward-opening parabola is plotted on the graph.

The parabola represents the graph of the quadratic function f(x)=2x2+6x+4.5
. Which of the following describes the domain of this function?

(1 point)
Responses

−1.5≤x≤0
negative 1.5 less-than-or-equal-to x less-than-or-equal-to 0

x≤0
x is less than or equal to 0

x≥0
x is greater than or equal to 0

−∞<x<∞

Graph the quadratic y=4x^2+3x−8
. What is the vertex of this quadratic?(1 point)
Responses

(−0.375,−8.563)
left parenthesis negative 0.375 comma negative 8.563 right parenthesis

(0,−8)
left parenthesis 0 comma negative 8 right parenthesis

(−1.838,0)
left parenthesis negative 1.838 comma 0 right parenthesis

(1.088,0)

Use the image to answer the question.

A parabola is graphed on a coordinate plane. The x-axis ranges from negative 10 to 3 in increments of 1. The y-axis ranges from negative 10 to 10 in increments of 1.

Determine whether the a-value is positive or negative. Provide an appropriate reason.

(1 point)
Responses

The a-value is negative because the x-intercepts are negative.
The a -value is negative because the x -intercepts are negative.

The a-value is positive because the graph is of a parabola that opens up.
The a -value is positive because the graph is of a parabola that opens up.

The a-value is negative because the y-intercept is negative.
The a -value is negative because the y -intercept is negative.

The a-value is negative because the graph is of a parabola that opens down.

Convert the equation from vertex form to standard form y=−4(x+5)2−3
.(1 point)
Responses

y=16x2+160x+397
y equals 16 x squared plus 160 x plus 397

y=−4x2−40x−103
y equals negative 4 x squared minus 40 x minus 103

y=16x2+160x+412
y equals 16 x squared plus 160 x plus 412

y=−4x2−25x−3

Answer 1

Let's break down each of the questions individually:

1. Determine the domain and range of the function f(x) = x² + 2x - 15:

   • Domain: The domain of a quadratic function is all real numbers, which can be expressed as \( R \) or \(-\infty < x < \infty\).
   • Range: To find the range, we first determine the vertex. The vertex form is obtained by completing the square or using the vertex formula \( x = -\frac{b}{2a} \). For \( f(x) = x^2 + 2x - 15 \), \( a = 1 \) and \( b = 2 \):
     • \( x = -\frac{2}{2 \cdot 1} = -1 \).
     • Now, find \( f(-1) \): \[ f(-1) = (-1)^2 + 2(-1) - 15 = 1 - 2 - 15 = -16. \]
     • This is the minimum point, so the range is \( y \geq -16 \).
   • Final response: domain R; range y ≥ -16.

2. Describe the domain of the function f(x) = 2x² + 6x + 4.5:

   • The function is a quadratic function, which means the domain is all real numbers.
   • Final response: −∞ < x < ∞.

3. Find the vertex of the quadratic y = 4x² + 3x - 8:

   • Using the vertex formula \( x = -\frac{b}{2a} \) where \( a = 4 \) and \( b = 3 \):
     • \( x = -\frac{3}{2 \cdot 4} = -\frac{3}{8} \).
   • Now substitute \( x = -\frac{3}{8} \) back into the equation to find \( y \): \[ y = 4\left(-\frac{3}{8}\right)^2 + 3\left(-\frac{3}{8}\right) - 8 \] \[ = 4\left(\frac{9}{64}\right) - \frac{9}{8} - 8 \] \[ = \frac{36}{64} - \frac{72}{64} - \frac{512}{64} = \frac{36 - 72 - 512}{64} = \frac{-548}{64} = -8.5625. \]
   • The vertex is approximately \( (-0.375, -8.5625) \).
   • Final response: (−0.375,−8.563).

4. Determine whether the a-value is positive or negative:

   • If the parabola opens upwards, the a-value is positive, and if it opens downwards, the a-value is negative.
   • As stated, the parabola opens upwards.
   • Final response: The a-value is positive because the graph is of a parabola that opens up.

5. Convert the equation from vertex form to standard form y = −4(x + 5)² − 3:

   • Start by expanding: \[ y = −4(x^2 + 10x + 25) - 3 \] \[ = -4x^2 - 40x - 100 - 3 \] \[ = -4x^2 - 40x - 103. \]
   • Final response: y = −4x² − 40x − 103.

Please let me know if you need further clarification on any of the points!