Question

question 7 alg.2: spiral review - operations of polynomials solve the equation by graphing. $x^2 - 12x + 27 = 0$ first, graph the associated parabola by plotting the vertex and four additional points, two on each side of the vertex. then, use the graph to give the solution(s) to the equation. if there is more than one solution, separate them with commas. solution(s): x =

Explanation:

Step1: Find the vertex of the parabola

For a quadratic function \( y = ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). For the equation \( y = x^{2}-12x + 27 \), \( a = 1 \), \( b=- 12 \), \( c = 27 \). So \( x=-\frac{-12}{2\times1}=6 \). Substitute \( x = 6 \) into the equation to find the y - coordinate: \( y=6^{2}-12\times6 + 27=36-72 + 27=-9 \). So the vertex is \( (6,-9) \).

Step2: Find two points on each side of the vertex

• For \( x = 4 \): \( y=4^{2}-12\times4 + 27=16-48 + 27=-5 \)
• For \( x = 5 \): \( y=5^{2}-12\times5 + 27=25-60 + 27=-8 \)
• For \( x = 7 \): \( y=7^{2}-12\times7 + 27=49-84 + 27=-8 \)
• For \( x = 8 \): \( y=8^{2}-12\times8 + 27=64-96 + 27=-5 \)

Step3: Graph the parabola and find the x - intercepts

The x - intercepts of the parabola \( y=x^{2}-12x + 27 \) are the solutions to the equation \( x^{2}-12x + 27 = 0 \). We can also factor the quadratic: \( x^{2}-12x + 27=(x - 3)(x - 9)=0 \). So the solutions are \( x = 3 \) and \( x = 9 \).

Answer:

3, 9