1-secA+tanA/1+secA-tanA = secA+tanA-1

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?

Find the gradient of f(x, y,z)equals=left parenthesis x squared plus y squared plus z squared right parenthesis superscript negative 1 divided by 2 baseline plus ln left parenthesis x right parenthesis x2+y2+z2−1/2+ln(xyz) at the point left parenthesis negative 2 comma 1 comma negative 2 right parenthesis(−2,1,−2).

[10 points, algebra 2-simplifying complex fractions]according to my answer key the restrictions are -1 and 4. but why not positive 1?