Finding the vertex of a quadratic function:5x2−20x+15

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Emelia Leon · Accepted Answer

Step 1Let&#039;s say you have a quadratic functionf(x)=ax2+bx+cThen, you create a new quadratic function by multiplying by a factor of λ:g(x)=λf(x)=λax2+λbx+λcPutting f into vertex form, we have:y=ax2+bx+cy=a(x2+bax)+cy=a(x2+bax+b24a2)−b24a+cy=a(x−(−b2a))2−b24a+cy−(c−b24a)=a(x−(−b2a))2Comparing this to vertex form, y−k=a(h−k)2, we have the following coordinates for the vertex of f:(hf,kf)=(−b2a,c−b24a).Now, let&#039;s consider what happens to the vertex if we scale our quadratic f by a factor of λ:g(x)=λf(x)=λax2+λbx+λc.Following the same process of completing the square, we have:y=λax2+λbx+λcy=λa(x2+bax)+λcy=λa(x2+bax+b24a2)−λb24a+λcy=λa(x−(−b2a))2−λb24a+λcy−λ(c−b24a)=λa(x−(−b2a))2y−λhf=λa(x−kf)2Notice, the vertex of g is at coordinates(hg,kg)=(λhf,kf).So, scaling our function by a factor of λ leaves the x-coordinate of the vertex unchanged, but it does scale the y-coordinate of the vertex by that same factor.In this case, that scaling factor was λ=15, so the x-coordinate was the same for both:hf=hg=2.But, the y-coordinate of the scaled function was 15 the y-coordinate of the original function:hg=−1=15×−5=15×hf.

Finding the vertex of a quadratic function: \(\displaystyle{5}{x}^{{2}}-{20}{x}+{15}\)

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