Question

Find the equation of the quadratic function that has vertex at (-3,6) and passes through the point (2,-11).

Answer 1

Answer:

The quadratic equation is $\displaystyle\bf y=-\frac{17}{25} x^2-\frac{102}{25} x-\frac{3}{25}$.

Step-by-step explanation:

To find the quadratic equation that has a vertex at (-3, 6) and passes through the point (2, -11), we assume that the quadratic equation is:

$\boxed{y=ax^2+bx+c}$

Given:

• vertex at (-3, 6) which means:

        (i) the axis of symmetry is x = -3

        (ii) the function passes through the point (-3, 6)

• the function passes through the point (2, -11)

Using these informations, we can create a system of equation:

(1) axis of symmetry is x = -3

  The formula of axis of symmetry: $\boxed{x=-\frac{b}{2a} }$

   $\displaystyle -\frac{b}{2a} =-3$

   $b=6a\ ...\ [1]$

(2) passes through the point (-3, 6)

    $y=ax^2+bx+c$

    $6=a(-3)^2+b(-3)+c$

    $9a-3b+c=6\ ...\ [2]$

(3) passes through the point (2, 11)

    $y=ax^2+bx+c$

    $11=a(2)^2+b(2)+c$

    $4a+2b+c=-11\ ...\ [3]$

Substituting [1] into [2]:

$9a-3b+c=6$

$9a-3(6a)+c=6$

$-9a+c=6\ ...\ [4]$

Substituting [1] into [3]:

$4a+2b+c=-11$

$4a+2(6a)+c=-11$

$16a+c=-11\ ...\ [5]$

Combining [4] & [5]

$-9a+c=6$

$16a+c=-11$

----------------------- (-)

   $-25a=17$

         $\displaystyle\bf a=-\frac{17}{25}$

Substitute the value of a to [1]

$b=6a$

$\displaystyle b=6\left(-\frac{17}{25} \right)$

$\bf\displaystyle b=-\frac{102}{25}$

Substitute the values of a and b to [3]

$4a+2b+c=-11$

$\displaystyle 4\left(-\frac{17}{25} \right)+2\left(-\frac{102}{25} \right)+c=-11$

$\displaystyle\bf c=-\frac{3}{25}$

Therefore, the quadratic equation is $\displaystyle\boxed{y=-\frac{17}{25} x^2-\frac{102}{25} x-\frac{3}{25}}$