SOLUTION: I am studying for my finals and I need help with
Consider y= -2x^2-8x+5
Express it in Vertex form, then find the vertex.
I figured part a. out, y=a(x-8)^2+5, xv= 8, yv= 5
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-> SOLUTION: I am studying for my finals and I need help with
Consider y= -2x^2-8x+5
Express it in Vertex form, then find the vertex.
I figured part a. out, y=a(x-8)^2+5, xv= 8, yv= 5
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Question 875255: I am studying for my finals and I need help with
Consider y= -2x^2-8x+5
Express it in Vertex form, then find the vertex.
I figured part a. out, y=a(x-8)^2+5, xv= 8, yv= 5
I still need help with b.
Is the vertex a maximum or minimum point? What is its value
You can put this solution on YOUR website! y = ax^2 + bx + c
y= -2x^2-8x+5 = -2(x+ )^2 + + 5
y = -2(x+2)^2 +13, V(-2,13) Opening Downward a<0, V(-2,13) max point
+
+ = -a(2)^2 = -(-2)(4) = 8
You can put this solution on YOUR website! The leading term in general form or the value of the coefficient on the squared binomial will tell you if the graph or equation has vertex as a minimum or as a maximum. y=ax^2+bx+c opens upward for a>0 and opens downward for a<0.
You should study more about how to complete the square to put a quadratic equation into standard form. The term to complete the square for your equation is 4; but you first need to factor the 2 from the two terms containing x.
You should find .
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