824635d7-d744-436c-8094-7ef525314fa3.html

<h4>Using Vertex Form</h4>
<p>The vertex form of a quadratic is below:</p>
<p><img height="157" src="https://k12.openstax.org/contents/raise/resources/1a98c55ddb5612a215745e1087e68a20afdfaa9a" width="550"></p>
<p>Notice that the value of \(h\) is the opposite of what it is in the parenthesis. This seems odd: when we add \(k\), the vertex of the parabola moves up \(k\) units. But when we subtract \(h\) within the squared term, the vertex of the parabola moves right \(h\) units, not left.</p>
<p>Let&rsquo;s explore why. We know the \(y\)-coordinate of the vertex is \(y = k\). To find that using this equation, we need to know where \(a(x-h)^2\) is zero, because that will make \(y=k\). Set \(a(x-h)^2=0\) . This happens when \(x-h=0\), which is when \(x=h\). That means \(h\) is the \(x\)-coordinate of the vertex.</p>
<p><strong>Example</strong></p>
<p>Find the vertex of \(y=3(x-4)^2-2\).</p>
<p>The parabola of the quadratic will shift right 4 (remember this horizontal shift is counterintuitive and translates in the opposite direction of what the expression &ldquo;looks like&rdquo;) and down 2.</p>
<p>So, the vertex is at \((4, -2)\).</p>
<h4>Try It: Using Vertex Form</h4>
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<div class="os-raise-ib-cta" data-button-text="Solution" data-fire-event="Reveal1" data-schema-version="1.0">
  <div class="os-raise-ib-cta-content">
    <p>Find the vertex of \(y= (x-2)^2+5\).</p>
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  <div class="os-raise-ib-cta-prompt">
    <p>Write down your answer. Then select the <strong>solution</strong> button to compare your work. </p>
  </div>
</div>
<div class="os-raise-ib-content" data-schema-version="1.0" data-wait-for-event="Reveal1">
  <p>Here is how to find the vertex:</p>
  <p>The quadratic shifts to the right 2 and up 5. The vertex is at \((2, 5)\).</p>
</div>