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UNO [17]
2 years ago
9

Consider h(x)=x^2+8+15. identify its vertex and y-intercept.

Mathematics
1 answer:
OleMash [197]2 years ago
5 0

Answer:

Vertex: (-4, -1)

Y-intercept: (0, 15)

Step-by-step explanation:

Given the quaratic function, h(x) = x² + 8x + 15:

In order to determine the vertex of the given function, we can use the formula, [x = \frac{-b}{2a}, h(\frac{-b}{2a})].

<h3>Use the equation:  [x = \frac{-b}{2a}, h(\frac{-b}{2a})]</h3>

In the quadratic function, h(x) = x² + 8x + 15, where:

a = 1, b = 8, and c = 15:

Substitute the given values for <em>a</em> and <em>b</em> into the equation to solve for the x-coordinate of the vertex.

x = \frac{-b}{2a}

x = \frac{-8}{2(1)}

x = -4

Subsitute the value of the x-coordinate into the given function to solve for the <u>y-coordinate of the vertex</u>:

h(x) = x² + 8x + 15

h(-4) = (-4)² + 8(-4) + 15

h(-4) = 16 - 32 + 15

h(-4) = -1

Therefore, the vertex of the given function is (-4, -1).

<h3>Solve for the Y-intercept:</h3>

The <u>y-intercept</u> is the point on the graph where it crosse the y-axis. In order to find the y-intercept of the function, set x = 0, and solve for the y-intercept:

h(x) = x² + 8x + 15

h(0) = (0)² + 8(0) + 15

h(0) = 0 + 0 + 15

h(0) = 15

Therefore, the y-intercept of the quadratic function is (0, 15).

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The number of visits to public libraries increased from 1.5 billion in 1991 to 1.7 billion in 1996. Find the average rate of cha
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The given parameters are:

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The average rate of change between 1991 and 1996 is calculated as:

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6 0
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Consider the following proability density function: f(x) = { kx. 0&lt;= x&lt; 2, k(4-x). 2&lt;= x &lt;=4, 0. otherwise. find the
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The value of K for which f(x) is a valid probability density function is 1/4.

<h3>How to solve for the value of K</h3>

\int\limits^4_0 {fx(x)} \, dx =1

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K\frac{4}{2}+K[8 - (\frac{16}{2}  -\frac{4}{2} )] = 1\\

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4 0
1 year ago
Someone please help
AleksandrR [38]

Replace y with 31:

31 = 7x -4

Add 4 to both sides:

35 = 7x

Divide both sides by 7:

x = 5

Input x = 5

6 0
2 years ago
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