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Question: How do you find the quadratic function with vertex \(\left( {5,12} \right)\) and point \(\left( {7,1...

How do you find the quadratic function with vertex (5,12)\left( {5,12} \right) and point (7,15)?\left( {7,15} \right)?

Explanation

Solution

We know that a quadratic function is a second degree polynomial equation of degree 22. The general form of the quadratic function is f(x)=ax2+bx+cf(x) = a{x^2} + bx + c. The general form of the quadratic function with vertex (h,k)\left( {h,k} \right) and as the constant multiplier is as follows: f(x)=a(xh)2+kf(x) = a{\left( {x - h} \right)^2} + k. We are going to substitute the values of the vertex and the point and then we get the required quadratic function.

Complete step by step solution:
As we know the general form of the quadratic function is f(x)=a(xh)2+kf(x) = a{\left( {x - h} \right)^2} + k. According to the question we have h=5h = 5 and k=12k = 12. By putting the values we have: f(x)=a(x5)2+12f(x) = a{(x - 5)^2} + 12.
Now by substituting the point (7,15)(7,15) in the above equation by replacing x=7x = 7 and f(x)=15f(x) = 15.
So the new equation is 15=a(75)2+1215 = a{(7 - 5)^2} + 12. We will now solve this equation, it gives us 15=a(2)2+1215=4a+1215 = a{(2)^2} + 12 \Rightarrow 15 = 4a + 12.
By isolating the term aa and transferring the constant to the same side: 4a=1512a=344a = 15 - 12 \Rightarrow a = \dfrac{3}{4}.
Now we will put the value of aa in the f(x)f(x) and the equation is f(x)=34(x5)2+12f(x) = \dfrac{3}{4}{(x - 5)^2} + 12.

Hence the quadratic function of the vertex form is f(x)=34(x5)2+12f(x) = \dfrac{3}{4}{(x - 5)^2} + 12.

Note: We should have proper knowledge of quadratic function, their vertex and points before solving this kind of question. The vertex form is an alternative way to write the equation of the parabola. We need to keep in mind that in vertex form , one should be careful about the positive and negative signs to avoid calculation mistakes.