NEED AN ANSWER RIGHT NOW I HAVE ALMOST NO TIME PLEASE The zeros of a quadratic relationship are 3 and 7. The optimal value is -3. Is the optimal value a maximum or minimum? Expl…

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asked Mar 5, 2023 88.2k views

1 Answer

Jinger · Answer 1 · 2023-03-12T08:08:00+0000

Answer:

Minimum

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:

$\sf x=(7-3)/(2)+3=5$

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

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:

$\sf 21a=25a-3$

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

$\sf \implies a=\frac34$

So the final equation is:

$\sf factor \ form \implies y=\frac34(x-3)(x-7)$

$\sf standard \ form\implies y=\frac34x^2-(15)/(2)x+(63)/(4)$

$\sf vertex \ form \implies y=\frac34(x-5)^2-3$