Question

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

1. x-int: (-2,0) , (5,0)
P (3,6)

Answer 1

Below

Explanation:

a(x-p)(x-q) p = -2 q = 5

a ( x- - 2)(x - 5) sub in the values of the point given (3,6) to calculate 'a'

6 = a (3+2)(3-5)

a = -6 /10 = - 3/5

y = -3/5 (x+2)(x-5)

Answer 2

`\textsf{Intercept form}: \quad y=-(3)/(5)(x+2)(x-5)`

`\textsf{Standard form}: \quad y=-(3)/(5)x^2+(9)/(5)x+6`

Explanation:

`\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}}`

If the x-intercepts are (-2, 0) and (5, 0) then:

• p = -2
• q = 5

Substitute the values of p and q into the formula:

`\implies y=a(x-(-2))(x-5)`

`\implies y=a(x+2)(x-5)`

To find a, substitute the given point on the curve P (3, 6) into the equation:

`\implies 6=a(3+2)(3-5)`

`\implies 6=a(5)(-2)`

`\implies 6=-10a`

`\implies a=(6)/(-10)`

`\implies a=-(3)/(5)`

Substitute the found value of a into the equation:

`\implies y=-(3)/(5)(x+2)(x-5)`

Expand to write the equation in standard form:

`\implies y=-(3)/(5)(x^2-3x-10)`

`\implies y=-(3)/(5)x^2+(9)/(5)x+6`