Mr. Walker gave his class the function f(x) = (x + 3)(x + 5). Four students made a claim about the function. Each student’s claim is below. Jeremiah: The y-intercept is at (15, 0). Lindsay: The x-intercepts are at (–3, 0) and (5, 0). Stephen: The vertex is at (–4, –1). Alexis: The midpoint between the x-intercepts is at (4, 0). Which student’s claim about the function is correct? The claim by is correct.

The [tex]y[/tex] intercept of second order polynomials is found for [tex]x = 0[/tex], [tex]x[/tex] intercepts are found for [tex]y = 0[/tex], vertex of second order polynomials are found by understanding polynomials in standard form.

y-Intercept

[tex]f(0) = (0+3)\cdot (0+5)[/tex]

[tex]f(0) = 15[/tex]

The [tex]y[/tex] intercept of the function is [tex](x,y) = (0, 15)[/tex]. (Jeremiah is wrong) [tex]\blacksquare[/tex]

x-Intercept

By direct inspection we find that [tex]x[/tex] intercepts are [tex](x,y) = (-3, 0)[/tex] and [tex](x,y) = (-5, 0)[/tex]. (Lindsay is wrong) [tex]\blacksquare[/tex]

Vertex

We expand the function, complete the square and rearrange the expression in standard form:

[tex]f(x) = (x+3)\cdot (x+5)[/tex]

[tex]f(x) = x^{2}+8\cdot x + 15[/tex]

[tex]f(x) + 1 = x^{2}+8\cdot x + 16[/tex]

[tex]y + 1 = (x+4)^{2}[/tex] (1)

The vertex of the parabola is [tex](x,y) = (-4, -1)[/tex]. (Stephen is right) [tex]\blacksquare[/tex]

Midpoint

The midpoint is found by following vectorial expression:

[tex](x,y) = 0.5\cdot (-3, 0) +0.5\cdot (-5,0)[/tex]

[tex](x,y) = (-4,0)[/tex]

The midpoint between the [tex]x[/tex] intercepts is [tex](-4,0)[/tex]. (Alexis is wrong) [tex]\blacksquare[/tex]

Stephen's claim about the function is correct. [tex]\blacksquare[/tex]

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