Answer:
Minimum
Step-by-step explanation:
The zeros of a quadratic equation are the points at which the parabola intersects the x-axis.
[tex]\sf x=3 \implies x-3=0[/tex]
[tex]\sf x=7 \implies x-7=0[/tex]
[tex]\sf \implies y=a(x-3)(x-7)[/tex] (for some constant a)
[tex]\sf \implies y=ax^2-10ax+21a[/tex]
The optimal value is the y-coordinate of the vertex.
[tex]\sf \implies vertex=(x,-3)[/tex]
The x-coordinate of the vertex is the midpoint of the zeros:
[tex]\sf x=\dfrac{7-3}{2}+3=5[/tex]
[tex]\sf \implies vertex=(5,-3)[/tex]
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: [tex]\sf y=a(x-h)^2+k[/tex]
where (h, k) is the vertex
[tex]\sf \implies y=a(x-5)^2-3[/tex]
[tex]\sf \implies y=ax^2-10ax+25a-3[/tex]
To find the value of a, compare the constants of both equations:
[tex]\sf 21a=25a-3[/tex]
[tex]\sf \implies -4a=-3[/tex]
[tex]\sf \implies a=\dfrac34[/tex]
So the final equation is:
[tex]\sf factor \ form \implies y=\dfrac34(x-3)(x-7)[/tex]
[tex]\sf standard \ form\implies y=\dfrac34x^2-\dfrac{15}{2}x+\dfrac{63}{4}[/tex]
[tex]\sf vertex \ form \implies y=\dfrac34(x-5)^2-3[/tex]
