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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

1 answer

Let's answer each question one by one based on the information provided.

  1. Domain and Range of f(x)=x²+2x−15:

    • Domain: Since this is a quadratic function, its domain is all real numbers \( \mathbb{R} \).
    • Range: To find the vertex, we can use the vertex formula \( x = -\frac{b}{2a} \) where \( a=1 \) and \( b=2 \): \[ x = -\frac{2}{2(1)} = -1 \] Substitute \( x = -1 \) into \( f(x) \): \[ f(-1) = (-1)² + 2(-1) - 15 = 1 - 2 - 15 = -16 \] Therefore, the minimum value of the function is \( -16 \). Thus, Range: \( y \geq -16 \).
    • Correct Response: domain R; range y ≥ −16

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

    • Since this is also a quadratic function, its domain is all real numbers \( -\infty < x < \infty \).
    • Correct Response: −∞<x<∞

  3. Vertex of the quadratic y=4x²+3x−8:

    • To find the vertex \( x = -\frac{b}{2a} \): Here, \( a = 4, b = 3 \): \[ x = -\frac{3}{2(4)} = -\frac{3}{8} \] Substitute \( x = -\frac{3}{8} \) into \( y \): \[ y = 4\left(-\frac{3}{8}\right)² + 3\left(-\frac{3}{8}\right) - 8 \] First, calculate \( (-\frac{3}{8})² = \frac{9}{64} \): \[ y = 4 \cdot \frac{9}{64} - \frac{9}{8} - 8 = \frac{36}{64} - \frac{72}{64} - \frac{512}{64} = -\frac{548}{64} = -8.563 \] Thus, the vertex is \( (-0.375, -8.563) \).
    • Correct Response: (−0.375,−8.563)

  4. Determination of the a-value for the parabola:

    • If the parabola opens upwards, the a-value is positive, and if it opens downwards, the a-value is negative. From your description that the parabola opens up:
    • Correct Response: The a-value is positive because the graph is of a parabola that opens up.

  5. Convert y=−4(x+5)²−3 to standard form: \[ y = -4(x² + 10x + 25) - 3 = -4x² - 40x - 100 - 3 = -4x² - 40x - 103 \]

    • Correct Response: y=−4x²−40x−103

Let me know if you have more questions!