heavenaubry
Answered

Find the equation of the parabola with the given x-intercepts and point on the graph. Use y = a(x-p)(x-q).

2. x-int: (-4,0) , (-10, 0)
P (-6,-12)



Answer :

mhanifa

Answer:

  • y = 1.5(x + 4)(x + 10)

=========================

Given

  • x- intercepts (-4, 0) and (-10, 0),
  • Point P (- 6, - 12).

Solution

The given translates as:

  • p = - 4, q = - 10, x = - 6, y = - 12

Use given x - intercepts to get the equation:

  • y = a(x + 4)(x + 10)

Use the coordinates of P to find the value of a:

  • - 12 = a(-6 + 4)(-6 + 10)
  • - 12 = a*(-2)*4
  • - 12 = - 8a
  • a = - 12 / - 8
  • a = 1.5

The equation of this parabola is:

  • y = 1.5(x + 4)(x + 10)
semsee45

Answer:

[tex]\textsf{Intercept form}: \quad y=\dfrac{3}{2}(x+4)(x+10)[/tex]

[tex]\textsf{Standard form}: \quad y=\dfrac{3}{2}x^2+21x+60[/tex]

Step-by-step explanation:

[tex]\boxed{\begin{minipage}{6 cm}\underline{Intercept form of a quadratic equation}\\\\$y=a(x-p)(x-q)$\\\\where:\\ \phantom{ww}$\bullet$ $p$ and $q$ are the $x$-intercepts. \\ \phantom{ww}$\bullet$ $a$ is some constant.\\\end{minipage}}[/tex]

If the x-intercepts are (-4, 0) and (-10, 0) then:

  • p = -4
  • q = -10

Substitute the values of p and q into the formula:

[tex]\implies y=a(x-(-4))(x-(-10))[/tex]

[tex]\implies y=a(x+4)(x+10)[/tex]

To find a, substitute the given point on the curve P (-6, -12) into the equation:

[tex]\implies -12=a(-6+4)(-6+10)[/tex]

[tex]\implies -12=a(-2)(4)[/tex]

[tex]\implies -12=-8a[/tex]

[tex]\implies a=\dfrac{-12}{-8}[/tex]

[tex]\implies a=\dfrac{3}{2}[/tex]

Substitute the found value of a into the equation:

[tex]\implies y=\dfrac{3}{2}(x+4)(x+10)[/tex]

Expand to write the equation in standard form:

[tex]\implies y=\dfrac{3}{2}(x^2+14x+40)[/tex]

[tex]\implies y=\dfrac{3}{2}x^2+21x+60[/tex]

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