Find the general form of the equation for the quadratic function using the vertex&nbsp;(h,&nbsp;k)&nbsp;and a point on the graph&nbsp;(x,&nbsp;y).&nbsp;(h,&nbsp;k)=(1,&nbsp;0)(x,&nbsp;y)=(0,&nbsp;1)

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Find the general form of the equation for the quadratic function using the vertex&amp;nbsp;(h,&amp;nbsp;k)&amp;nbsp;and a point on the graph&amp;nbsp;(x,&amp;nbsp;y).&amp;nbsp;(h,&amp;nbsp;k)=(1,&amp;nbsp;0)(x,&amp;nbsp;y)=(0,&amp;nbsp;1)

David Tyson · Accepted Answer

Step 1 The general form of parabula is: y=a(x−h)2+k Given that (h, k)=(1, 0) So, y=a(x−1)2+0 ⇒y=a(x−1)2 Now, the curve puos through (0, 1) 1=a(0−1)2 ⇒a=1 Hence, the required equation y=1(x−1)2+0 ⇒y=(x−1)2 So, f(x)=(x−1)2

Don Sumner · Accepted Answer

Answer:y=(x−1)2Explanation:y=a(x−h)2+kwhere (h, k) represents the vertex of the parabola.Given that the vertex is (1, 0) and a point on the graph is (0, 1), we can substitute these values into the equation to solve for &#039;a&#039;.Using the vertex coordinates, we have:0=a(1−1)2+0Simplifying the equation, we get:0=a(0)+00=0Since the equation simplifies to 0 = 0, this equation does not provide any information about &#039;a&#039;. Therefore, we need to use the point (0, 1) to solve for &#039;a&#039;.Substituting the point coordinates into the equation, we have:1=a(0−1)2+0Simplifying further:1=a(−1)21=a(1)1=aNow that we have found the value of &#039;a&#039; as 1, we can substitute it back into the vertex form equation to obtain the general form of the quadratic function:y=1(x−1)2+0Simplifying:y=(x−1)2Thus, the general form of the quadratic function is y=(x−1)2.

nick1337 · Accepted Answer

Given:y=a(x−h)2+k.Given that the vertex is (h,k)=(1,0) and a point on the graph is (x,y)=(0,1), we can substitute these values into the equation:1=a(0−1)2+0.Simplifying the equation, we have:1=a(−1)2.Squaring −1, we get:1=a.Therefore, the general form of the equation for the quadratic function is:y=1(x−1)2+0.

RizerMix · Accepted Answer

Step 1:To find the general form of the equation for a quadratic function using the vertex (h,k) and a point on the graph (x,y), we can start with the vertex form of the quadratic equation:f(x)=a(x−h)2+kwhere (h,k) represents the vertex of the parabola. Plugging in the given vertex values (h,k)=(1,0), we have:f(x)=a(x−1)2+0Simplifying further, we get:f(x)=a(x−1)2Step 2:Now, let&#039;s substitute the coordinates of the point (x,y)=(0,1) into the equation:1=a(0−1)2Simplifying the expression on the right-hand side, we have:1=a(−1)2=aTherefore, the value of a is 1.Substituting this value back into the equation, we get the final general form of the quadratic function:f(x)=(x−1)2

Use the vertex (h,\ k) and a point on the

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