Algebra Know-It-ALL

Part Three 511

Question 24-4

Imagine another quadratic function in which x is the independent variable and y is the depen-
dent variable. If this function has a single real zero with multiplicity 2 where x=p, what is xv,
thex-value of the vertex point on its graph?

Answer 24-4

When a quadratic function has only one real zero, the parabola is tangent to the x axis at the
vertex point. That’s also the x-value of the real zero. Therefore, xv=p.

Question 24-5

Suppose we come across the following quadratic function in binomial factor form, where x is
the independent variable and y is the dependent variable:

y= (x+ 2)(x− 4)

Does the parabola representing this function in Cartesian coordinates open upward or down-
ward?

Answer 24-5

To determine this, we must get the right side of the equation in polynomial standard form by
multiplying the binomials. When we do that, we get

y=x^2 − 2 x− 8

Because the coefficient of x^2 is positive, the parabola opens upward.

Question 24-6

What are the real zeros of the function stated in Question 24-5? What are the coordinates (xv,yv)
of the vertex point in its graph? Is the vertex an absolute maximum or an absolute minimum?

Answer 24-6

The zeros can be seen by looking at the original form of the function. The right side of that equa-
tion is a product of binomials. If we set it equal to 0, getting a quadratic equation in x, we have

(x+ 2)(x− 4) = 0

The zeros of the function are the same as the roots of this quadratic. Without doing any alge-
bra or arithmetic, we can see that these roots are x=−2 or x= 4.
To find the vertex point, let’s remember the general polynomial standard form for a qua-
dratic function:

y=ax^2 +bx+c

Thex-coordinate of the vertex point, xv, can be found by the formula

xv=−b/2a