Answer:
[tex]\textsf{Intercept form}: \quad y=-\dfrac{3}{5}(x+2)(x-5)[/tex]
[tex]\textsf{Standard form}: \quad y=-\dfrac{3}{5}x^2+\dfrac{9}{5}x+6[/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 (-2, 0) and (5, 0) then:
Substitute the values of p and q into the formula:
[tex]\implies y=a(x-(-2))(x-5)[/tex]
[tex]\implies y=a(x+2)(x-5)[/tex]
To find a, substitute the given point on the curve P (3, 6) into the equation:
[tex]\implies 6=a(3+2)(3-5)[/tex]
[tex]\implies 6=a(5)(-2)[/tex]
[tex]\implies 6=-10a[/tex]
[tex]\implies a=\dfrac{6}{-10}[/tex]
[tex]\implies a=-\dfrac{3}{5}[/tex]
Substitute the found value of a into the equation:
[tex]\implies y=-\dfrac{3}{5}(x+2)(x-5)[/tex]
Expand to write the equation in standard form:
[tex]\implies y=-\dfrac{3}{5}(x^2-3x-10)[/tex]
[tex]\implies y=-\dfrac{3}{5}x^2+\dfrac{9}{5}x+6[/tex]
