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3 years ago

Create a system of equations and use algebra to write a quadratic function that passes through the

Mathematics

1 answer:

skad [1K]3 years ago

6 0

y = x² + 3x + 10

the equation of a parabola in standard form is → y = ax² + bx + c ( a ≠ 0 )

Substitute each of the pairs of coordinate points into the equation and solve for a, b and c

(- 2, 8) → 8 = 4a + 2b + c → (1)

(1, 14) → 14 = a + b + c → (2)

note that (0 , 10) is the y-intercept ⇒ c = 10

substitute c = 10 into (1) and (2)

8 = 4a - 2b + 10 → 4a - 2b = - 2 → (3)

14 = a + b + 10 → a + b = 4 → (4)

multiply (4 ) by 2

2a + 2b = 8 → (5)

add (3) and (5) term by term ⇒ 6a = 6 ⇒ a = 1

substitute a = 1 into (4)

1 + b = 4 ⇒ b = 3

thus a = 1, b= 3 and c = 10

equation of parabola is y = x² + 3x + 10

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Mrs.Carle returned 10 books to the library. When she looked back in her classroom,she now had 88 books. Write an algebraic equat

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A line includes the points (4, 8) and (3, 5). What is its equation in slope-intercept form? Write your answer using integers, pr

stich3 [128]

Answer:

A line includes the points (4, 8) and (3, 5) whose equation in slope-intercept form will be :

$y=3x-4$

Step-by-step explanation:

The point slope form: $(x_1,y_1) and(x_2,y_2)$

$(y-y_1)=m(x-x_1)$

where : m = Slope of the line

$m=\frac{y_2-y_1}{x_2-x_1}$

A line includes the points (4, 8) and (3, 5).The equation of the line will be:

$m=\frac{5-8}{3-4}=3$

$(y-8)=3\times (x-4)$

$y-8=3x-12$

$y=3x-4$

(y=mx+c), where c is intercept at y axis.

The above equation represents slope-intercept form of the line.

6 0

3 years ago

The function f(x) = 1/3x + 5 and g(x) = x^2 + 4x are shown graphed below. The negative solution to the equation f(x) = g(x) is c

sergij07 [2.7K]

Answer:

Step-by-step explanation:

4 0

3 years ago

Find the six trig function values of the angle 240*Show all work, do not use calculator

-BARSIC- [3]

Solution:

Given:

$240^0$

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

$sin240^0=sin(180+60)$

Using the trigonometric identity;

$sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny$

Hence,

$\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}$

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

$cos240^0=cos(180+60)$

Using the trigonometric identity;

$cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny$

Hence,

$\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\ \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}$

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

$tan240^0=tan(180+60)$

Using the trigonometric identity;

$tan(180+x)=tan\text{ }x$

Hence,

$\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\ \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}$

To get cosec 240 degrees:

$\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}$

To get sec 240 degrees:

$\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\ \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\ \\ Thus, \\ sec240^0=-2 \end{gathered}$

To get cot 240 degrees:

$\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\ \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}$

5 0

1 year ago

Here 1 more than BIG PRIZE

Firlakuza [10]

Hey there! :D

Use the distributive property.

a(b+c)= ab+ac

6(9x+2)+2x

54x+12+2x

56x+ 12 <== equivalent expression

I hope this helps!
~kaikers

4 0

3 years ago