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Find the largest real number $c$ such that $1$ is in the range of $f(x)=x^2-5x+c-3x+8$.

 Jan 26, 2025

Best Answer 

 #1
avatar+14 
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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. 

 Jan 26, 2025
 #1
avatar+14 
+1
Best Answer

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. 

Kromy Jan 26, 2025

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