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fiasKO [112]
3 years ago
13

Write an equation of the line that passes through the given points. (-4,9) and (1, -1)

Mathematics
1 answer:
UkoKoshka [18]3 years ago
7 0

Answer:

y=-2x+1

Step-by-step explanation:

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What is the value of x in this equation
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Answer:

\textsf{A)} \quad x=-2, \:\:x=\dfrac{5}{2}

\textsf{B)} \quad \left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)

C)  See attachment.

Step-by-step explanation:

Given function:

f(x)=2x^2-x-10

<h3><u>Part A</u></h3>

To factor a <u>quadratic</u> in the form  ax^2+bx+c<em> , </em>find two numbers that multiply to ac and sum to b :

\implies ac=2 \cdot -10=-20

\implies b=-1

Therefore, the two numbers are -5 and 4.

Rewrite b as the sum of these two numbers:

\implies f(x)=2x^2-5x+4x-10

Factor the first two terms and the last two terms separately:

\implies f(x)=x(2x-5)+2(2x-5)

Factor out the common term  (2x - 5):

\implies f(x)=(x+2)(2x-5)

The x-intercepts are when the curve crosses the x-axis, so when y = 0:

\implies (x+2)(2x-5)=0

Therefore:

\implies (x+2)=0 \implies x=-2

\implies (2x-5)=0 \implies x=\dfrac{5}{2}

So the x-intercepts are:

x=-2, \:\:x=\dfrac{5}{2}

<h3><u>Part B</u></h3>

The x-value of the vertex is:

\implies x=\dfrac{-b}{2a}

Therefore, the x-value of the vertex of the given function is:

\implies x=\dfrac{-(-1)}{2(2)}=\dfrac{1}{4}

To find the y-value of the vertex, substitute the found value of x into the function:

\implies f\left(\dfrac{1}{4}\right)=2\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)-10=-\dfrac{81}{8}

Therefore, the vertex of the function is:

\left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)

<h3><u>Part C</u></h3>

Plot the x-intercepts found in Part A.

Plot the vertex found in Part B.

As the <u>leading coefficient</u> of the function is positive, the parabola will open upwards.  This is confirmed as the vertex is a minimum point.

The axis of symmetry is the <u>x-value</u> of the <u>vertex</u>.  Draw a line at x = ¹/₄ and use this to ensure the drawing of the parabola is <u>symmetrical</u>.

Draw a upwards opening parabola that has a minimum point at the vertex and that passes through the x-intercepts (see attachment).

5 0
2 years ago
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