Answer:
[tex]y=4x^2+5x-6[/tex]
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 [tex]y= ax^2+bx+c[/tex]
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 [tex]y=4x^2+5x-6[/tex]
we can see this in the graph attached.
The equation of the parabola is [tex]\bf y=4{x^2}+5x-6[/tex].
Further explanation:
The equation of parabola in form of quadratic equation is as follows:
[tex]y=a{x^2}+bx+c[/tex]
The given points are shown below in Table 1.
Step 1:
The parabola passes through point [tex]\left({-2,0}\right)[/tex], it can be written as follows:
[tex]\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}[/tex]
Step 2:
The parabola passes through point [tex]\left({0,-6}\right)[/tex], it can be written as follows:
[tex]\begin{aligned}-6&=a{\left(0\right)^2}+b\left(0\right)+c\hfill\\c&=-6\hfill\\\end{aligned}[/tex]
Step 3:
The parabola passes through point [tex]\left({4,78}\right)[/tex], it can be written as follows:
[tex]\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}[/tex]
From step 2, substitute [tex]-6[/tex] for [tex]c[/tex] in equation (1).
[tex]\begin{aligned}4a-2b-6&=0\,\,\\4a-2b&=6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left(3\right)\\ \end{aligned}[/tex]
From step 2, substitute -6 for [tex]c[/tex] in equation (2).
[tex]\begin{aligned}16a+4b-6&=78\\ 16a+4b&=84\\4a+b&=21\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left(4\right)\\\end{aligned}[/tex]
Subtract equation (3) from equation (4).
[tex]\begin{aligned}b-\left({-2b}\right)&=21-6\\b+2b&=15\\3b&= 15\\b&=5\\\end{aligned}[/tex]
Substitute 5 for [tex]b[/tex] in equation (3) to obtain the value of [tex]{a}[/tex].
[tex]\begin{aligned}4a-2\left(5\right)=6\\4a-10=6\\4a=16\\a=4\\\end{aligned}[/tex]
Substitute 4 for [tex]{a}[/tex], 5 for [tex]b[/tex] and [tex]-6[/tex] for [tex]c[/tex] in equation [tex]y=a{x^2}+bx+c[/tex] to obtain the equation of the parabola.
[tex]y=4{x^2}+5x-6[/tex]
The graph of the parabola is shown below in figure 1.
Thus, the equation of the parabola is [tex]\bf y=4{x^2}+5x-6[/tex].
Learn more:
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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
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3. what are the domain and range of the function f(x) = 3x + 5?
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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.
