Question

Which function has a vertex at (2, -9)?

Answer 1

The function $f(x)=\left(x-5\right)\left(x+1\right)$ has a vertex at (2, -9).

Quadratic function

The quadratic function can be represented by a quadratic equation in the Standard form: $ax^2+bx+c$ where: a, b and c are your respective coefficients. In the quadratic function the coefficient "a" must be different than zero ($a\neq 0$) and the degree of the function must be equal to 2.

For solving this question, you should find the vertex of quadratic equation (h,k). Where, $h=-\frac{b}{2a}$ . On the other hand, the value of the k-coordinate will be found when you replace the value of h in the equation of your question.

     1.  $f(x)=-(x-3)^2$

        Write this equation in Standard form:

                          $f(x)=-(x^2-6x+9)\\ \\ f(x)=-x^2+6x-9$

        Find h, using the formula  $h=-\frac{b}{2a}$

             $h=-\frac{b}{2a} =-\frac{6}{2*(-1)} =-\frac{6}{-2}=-(-3)=3$

        Find k, replacing x for the value of h

         $-x^2+6x-9\\ \\ -(3)^2+6*3-9=-9+18-9=-18+18=0$

      The vertex of quadratic equation is (3,0). Incorrect Option.

     2.  $f(x)=(x+8)^2$

        Write this equation in Standard form:

                          $f(x)=x^2+16x+64$

        Find h, using the formula  $h=-\frac{b}{2a}$

             $h=-\frac{b}{2a} =-\frac{16}{2*1} =-\frac{16}{2}=-8$

        Find k, replacing x for the value of h

         

         $x^2+16x+64\\ \\ (-8)^2+16*(-8)+64=64-128+64=128-128=0$

  The vertex of quadratic equation is (-8,0). Incorrect Option.

   3.  $f(x)=\left(x-5\right)\left(x+1\right)$

        Write this equation in Standard form:

                          $f(x)=(x^2+x-5x-5)\\ \\ f(x)=(x^2-4x-5)$

        Find h, using the formula  $h=-\frac{b}{2a}$

             $h=-\frac{b}{2a} =-\frac{-4}{2*1} =-\frac{-4}{2}=-(-2)=2$

        Find k, replacing x for the value of h

         $x^2-4x-5\\ \\ 2^2-4*2-5=4-8-5=-4-5=-9$

   The vertex of quadratic equation is (2,-9). Correct Option.

   4.  $f(x)=-\left(x-1\right)\left(x-5\right)$

        Write this equation in Standard form:

                          $f(x)=-\left(x-1\right)\left(x-5\right)\\ \\ f(x)=-(x^2-5x-x+5)\\ \\ f(x)=-(x^2-6x+5)\\ \\ f(x)=-x^2+6x-5$

        Find h, using the formula  $h=-\frac{b}{2a}$

             $h=-\frac{b}{2a} =-\frac{6}{2*(-1)} =-\frac{6}{-2}=-(-3)=3$

        Find k, replacing x for the value of h

         $-x^2+6x-5\\ \\ -(3)^2+6*3-5=-9+18-5=18-14=4$

The vertex of quadratic equation is (3,4). Incorrect Option.

Learn more about the vertex of the quadratic function here:

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