Find the largest real number $c$ such that $1$ is in the range of $f(x)=x^2-5x+c-3x+8$.
Rewrite the quadratic as \( x^2 -8x + (8 + c)\). This quadratic forms a "U-shape." Thus, we want the vertex to have a y coordinate of 1. The vertex occurs at \(x =\frac{-b}{2a}\) for the quadratic ax^2 + bx + c. Plugging in our values, we have \(\frac{8}{2} = 4\). This is the x-coordinate to the vertex of the parabola. Plugging in this number and setting the equation equal to 1, we have 16 - 32 + 8 + C = 1. Therefore, C = 9 generates a vertex of (4,1). Any higher C would generate a vertex with a y coordinate greater than 1. Thus the largest real number C would be 9.
Rewrite the quadratic as \( x^2 -8x + (8 + c)\). This quadratic forms a "U-shape." Thus, we want the vertex to have a y coordinate of 1. The vertex occurs at \(x =\frac{-b}{2a}\) for the quadratic ax^2 + bx + c. Plugging in our values, we have \(\frac{8}{2} = 4\). This is the x-coordinate to the vertex of the parabola. Plugging in this number and setting the equation equal to 1, we have 16 - 32 + 8 + C = 1. Therefore, C = 9 generates a vertex of (4,1). Any higher C would generate a vertex with a y coordinate greater than 1. Thus the largest real number C would be 9.