Question

what is the equation in standard form of a parabola that models the values in the table x -2 0 4 f(x) 0 -6 78

Answer 1

The equation of the parabola is $\bf y=4{x^2}+5x-6$.

Further explanation:

The equation of parabola in form of quadratic equation is as follows:

$y=a{x^2}+bx+c$

The given points are shown below in Table 1.

Step 1:

The parabola passes through point $\left({-2,0}\right)$, it can be written as follows:

$\begin{aligned}0&=a{\left({-2}\right)^2}+b\left({-2}\right)+c\hfill\\0&=4a-2b+c\hfill\\4a-2b+c&=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left(1 \right)\hfill\\ \end{aligned}$

Step 2:

The parabola passes through point $\left({0,-6}\right)$, it can be written as follows:

$\begin{aligned}-6&=a{\left(0\right)^2}+b\left(0\right)+c\hfill\\c&=-6\hfill\\\end{aligned}$

Step 3:

The parabola passes through point $\left({4,78}\right)$, it can be written as follows:

$\begin{aligned}78&=a{\left(4\right)^2}+b\left(4\right)+c\hfill\\78&=16a+4b+c\hfill\\16a+4b+c &=78\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left(2\right)\hfill\\ \end{aligned}$

From step 2, substitute $-6$ for $c$ in equation (1).

$\begin{aligned}4a-2b-6&=0\,\,\\4a-2b&=6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left(3\right)\\ \end{aligned}$

From step 2, substitute -6 for $c$  in equation (2).

$\begin{aligned}16a+4b-6&=78\\ 16a+4b&=84\\4a+b&=21\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left(4\right)\\\end{aligned}$

Subtract equation (3) from equation (4).

$\begin{aligned}b-\left({-2b}\right)&=21-6\\b+2b&=15\\3b&= 15\\b&=5\\\end{aligned}$

Substitute 5 for $b$ in equation (3) to obtain the value of ${a}$.

$\begin{aligned}4a-2\left(5\right)=6\\4a-10=6\\4a=16\\a=4\\\end{aligned}$

Substitute 4 for ${a}$, 5 for $b$ and $-6$ for $c$ in equation $y=a{x^2}+bx+c$ to obtain the equation of the parabola.

$y=4{x^2}+5x-6$

The graph of the parabola is shown below in figure 1.

Thus, the equation of the parabola is $\bf y=4{x^2}+5x-6$.

Learn more:  

1. Which function has an inverse that is also a function? {(–1, –2), (0, 4), (1, 3), (5, 14), (7, 4)} {(–1, 2), (0, 4), (1, 5), (5, 4), (7, 2)} {(–1, 3), (0, 4), (1, 14), (5, 6), (7, 2)} {(–1, 4), (0, 4), (1, 2), (5, 3), (7, 1)}  

brainly.com/question/1632445  

2. A given line has the equation 10x + 2y = −2. what is the equation, in slope-intercept form, of the line that is parallel to the given line and passes through the point (0, 12)? y = ( )x + 12  

brainly.com/question/1473992  

3. what are the domain and range of the function f(x) = 3x + 5?  

brainly.com/question/3412497  

Answer Details :

Grade: High School.  

Subject: Mathematics.  

Chapter: Coordinate geometry.  

Keywords:

Parabola, standard form of the parabola, y=ax^2+bx+c, quadratic equation, vertex of the parabola, y=4x^2+5x-6, intervals, intercepts, function value, intercepts of lines, slope, slope intercept form, continuous, range, point.

Answer 2

Answer:

 $y=4x^2+5x-6$

Step-by-step explanation:

Given points are :

  x     =    -2      0       4

y=f(x) =    0       -6      78

We have to model the parabola with the help of these points

Solution:

We consider a standard equation of parabola $y= ax^2+bx+c$

now, we put the points in the equation we get,

at (-2,0) is    4a-2b+c=0

at (0,-6) is    c=-6

at (4,78) is    78= 16a+4b+c  

now, solving these equation we get, a= 4    , b= 5    , c= -6

so the equation formed with these points is   $y=4x^2+5x-6$

we can see this in the graph attached.