Question

NEED AN ANSWER RIGHT NOW I HAVE ALMOST NO TIME PLEASE

Answer 1

Answer:

Minimum

Step-by-step explanation:

The zeros of a quadratic equation are the points at which the parabola intersects the x-axis.

$\sf x=3 \implies x-3=0$

$\sf x=7 \implies x-7=0$

$\sf \implies y=a(x-3)(x-7)$   (for some constant a)

$\sf \implies y=ax^2-10ax+21a$

The optimal value is the y-coordinate of the vertex.  

$\sf \implies vertex=(x,-3)$

The x-coordinate of the vertex is the midpoint of the zeros:

$\sf x=\dfrac{7-3}{2}+3=5$

$\sf \implies vertex=(5,-3)$

Therefore, the vertex will be in Quadrant IV and so the parabola opens upwards into Quadrant I.  

So the optimal value is a MINIMUM since the vertex is the minimum point of the curve.

Additional Information to create the equation of the quadratic

Vertex form of quadratic equation:  $\sf y=a(x-h)^2+k$

where (h, k) is the vertex

$\sf \implies y=a(x-5)^2-3$

$\sf \implies y=ax^2-10ax+25a-3$

To find the value of a, compare the constants of both equations:

$\sf 21a=25a-3$

$\sf \implies -4a=-3$

$\sf \implies a=\dfrac34$

So the final equation is:

$\sf factor \ form \implies y=\dfrac34(x-3)(x-7)$

$\sf standard \ form\implies y=\dfrac34x^2-\dfrac{15}{2}x+\dfrac{63}{4}$

$\sf vertex \ form \implies y=\dfrac34(x-5)^2-3$