The function f is defined by f(x)=x^2+3x-10 if f(x+5)=x^2+kx+30, k= find the smallest zero of f(x+5), x=

The function f is defined by f(x)=x^2+3x-10 If f(x+5)=x^2+kx+30, k= Find the smallest zero of f(x+5), x= By definition, the zero of a polynomial function is a number a such that f(a)=0. For quadratic equation, the zeros can be found on equalizing f to 0. First, f(x+5)=(x+5)²+3(x+5)-10=x²+10x+25+3x+15-10=x²+13x+30, and it is told that If f(x+5)=x^2+kx+30 = x²+13x+30, it implies k=13, so let’s find the zeros. x²+13x+30=0 is to solve: D= 13²-4x1x30=169-120=49, sqrt49=7, so x=-13-7 / 2=-20/2 = -10, and x= -13+7/ 2 = -6 /2= -3. Obviously, we know that -10< -3, so the smallest zero is -10.

It equals 1373.5 that’s the solution