Question:
Write an equation (any form) for the quadratic graphed below.
A quadratic equation is of the form
{eq}y = ax^{2} + bx + c {/eq}
where {eq}a {/eq}, {eq}b {/eq} and {eq}c {/eq} are constants and {eq}a \neq 0 {/eq}. A quadratic equation has two roots, meaning that it has two points at which the value of the function becomes zero. The two roots can be the same to different values. A quadratic equation also has a local maximum or minimum at which the derivative of the function with respect to {eq}x {/eq} becomes zero.
Answer and Explanation: 1
Let the general form of the quadratic equation be:
{eq}y = ax^{2} + bx + c \text{ - - - - - - - - - - - - - (1)} {/eq}
where {eq}a {/eq}, {eq}b {/eq} and {eq}c {/eq} are constants.
From the graph, we can see that the function passes through the point {eq}(1,1) {/eq}. It means that when {eq}x=1, y=1 {/eq}. Substituting this in equation (1), we get,
{eq}1 = a(1)^{2} + b\times 1 + c\\ \implies a + b + c + = 1\\ \implies a + b + c = 1 \text{ - - - - - - - - - - - - - (2)} {/eq}
As seen in the graph, the function also passes through {eq}(3,3) {/eq}. Applying this in equation (1),
{eq}3 = a\times (3)^{2} + b\times 3 + c\\ \implies 9a + 3b + c = 3 \text{ - - - - - - - - - - - - - (3)} {/eq}
From the graph, we can see that the local maximum of the quadratic function is at {eq}x = 3 {/eq}. To get the local maximum, we differentiate the function with respect to {eq}x {/eq} and put {eq}\dfrac{d}{dx}y = 0 {/eq}
Differentiating equation (1) with respect to {eq}x {/eq} we get,
{eq}\dfrac{d}{dx}y = a\dfrac{d}{dx}(x^{2}) + b\dfrac{d}{dx}x + \dfrac{d}{dx}c = 0\\ \implies a\times 2x + b + 0 = 0\\ \implies 2ax + b = 0 \text{ - - - - - - - - - - - - - (4)} {/eq}
Since the maximum is at {eq}x = 3 {/eq} substitute this in equation (4)
{eq}2\times 3\times a + b = 0\\ \implies 6a + b = 0 \text{ - - - - - - - - - - - - - (5)} {/eq}
Now we need to solve the two equations (2), (3), and (5) in order to get the value of the constants.
To do that we subtract equation (2) from equation (3),
{eq}(3)-(2)\implies (9a-a) + (3b-b) + (c-c) = (3-1)\\ \implies 8a + 2b = 2\\ \implies 4a + b = 1 \text{ - - - - - - - - - - - - - (6)} {/eq}
We then subtract equation (6) from equation (5).
{eq}(5) - (6)\implies (6a - 4a) + (b-b) = 1\\ \implies 2a = 1\\ \implies a = \dfrac{1}{2} {/eq}
To get {eq}b {/eq}, we substitute the value of {eq}a {/eq} in equation (5).
{eq}6a + b = 0\\ \implies 6\times \dfrac{1}{2} + b = 0\\ \implies 3 + b = 0\\ \implies b = -3 {/eq}
Now we have both {eq}a {/eq} and {eq}b {/eq}. Use these values in equation (2) to get the value of {eq}c {/eq}
{eq}a + b + c = 1\\ \implies \dfrac{1}{2} - 3 + c = 1\\ \implies c = \dfrac{1}{2} + 3 = \dfrac{7}{2} {/eq}
Now the equation for the quadratic function can be written by substituting the values of constants {eq}a {/eq}, {eq}b {/eq} and {eq}c {/eq} in equation (1).
{eq}y = \dfrac{1}{2}x^{2} - 3x + \dfrac{7}{2} {/eq}
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