The given quadratic function is
f(x)=x2+14x−29 f(x) = x^2 + 14x - 29 f(x)=x2+14x−29
From this, we identify the coefficients:
a=1,b=14,c=−29 a = 1, \quad b = 14, \quad c = -29 a=1,b=14,c=−29
To find the x-coordinate of the vertex h h h, we use the formula
h=−b2a h = -\frac{b}{2a} h=−2ab
Substituting the values of b b b and a a a:
h=−142⋅1=−7.0 h = -\frac{14}{2 \cdot 1} = -7.0 h=−2⋅114=−7.0
Next, we evaluate the original function at x=h x = h x=h to find the y-coordinate k k k:
k=f(h)=f(−7)=1⋅(−7)2+14⋅(−7)−29 k = f(h) = f(-7) = 1 \cdot (-7)^2 + 14 \cdot (-7) - 29 k=f(h)=f(−7)=1⋅(−7)2+14⋅(−7)−29
Calculating this step-by-step:
=49−98−29=−78.0 = 49 - 98 - 29 = -78.0 =49−98−29=−78.0
Now that we have h h h and k k k, we can express the function in vertex form:
f(x)=(x−h)2+k=(x−(−7.0))2−78.0 f(x) = (x - h)^2 + k = (x - (-7.0))^2 - 78.0 f(x)=(x−h)2+k=(x−(−7.0))2−78.0
This simplifies to:
f(x)=(x+7.0)2−78.0 f(x) = (x + 7.0)^2 - 78.0 f(x)=(x+7.0)2−78.0
The quadratic function in vertex form is
f(x)=(x+7)2−78 \boxed{f(x) = (x + 7)^2 - 78} f(x)=(x+7)2−78
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