# SOLUTION: Write each function in vertex form. Sketch the graph of the function and label its vertex. 33. y = x2 + 4x - 7 34. y = -x2 + 4x - 1 35. y = 3x2 + 18x 36.y = 1/2x2 - 5x +

Algebra ->  Algebra  -> College  -> Linear Algebra -> SOLUTION: Write each function in vertex form. Sketch the graph of the function and label its vertex. 33. y = x2 + 4x - 7 34. y = -x2 + 4x - 1 35. y = 3x2 + 18x 36.y = 1/2x2 - 5x +       Log On

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 Click here to see ALL problems on Linear Algebra Question 177856: Write each function in vertex form. Sketch the graph of the function and label its vertex. 33. y = x2 + 4x - 7 34. y = -x2 + 4x - 1 35. y = 3x2 + 18x 36.y = 1/2x2 - 5x + 12Found 2 solutions by stanbon, jojo14344:Answer by stanbon(57282)   (Show Source): You can put this solution on YOUR website!Write each function in vertex form. Sketch the graph of the function and label its vertex. 33. y = x^2 + 4x - 7 x^2 + 4x + ? = y+7+? x^2 + 4x + 4 = y+7+4 (x+2)^2 = y + 11 ----------------- Vertex: (-2,-11) ================================================== 34. y = -x^2 + 4x - 1 35. y = 3x^2 + 18x ------------------------- 36.y = 1/2x^2 - 5x + 12 (1/2)x^2 - 5x + ? = y-12 + ? (1/2)[x^2 - 10x + ? = y-12 + ? (1/2)[x^2 - 10x + 25] = y - 12 + (1/2)*25 (1/2)[x-5]^2 = y + (1/2) ---- Vertex: (5,-1/2)) ============================ Cheers, Stan H. Answer by jojo14344(1512)   (Show Source): You can put this solution on YOUR website! We know the standard eqn of a parabola: Being the vertex form---->, where (h,k) is the vertex: I'll do the first one, and you can continue the rest; 33. --->follows std eqn, where Complete the square, adding a constant by taking half of the "b" constant then squared. In this case the "b" constant is , half of it then squared, : , Also subtract what you added so the process won't change. ---->it follows the vertex form, being To get the x-intercept, we solve the eqn by Quadratic: where----> For the Y-Intercept, let As we see the graph: Thank you, Jojo