Question

How do you find the vertex and intercepts for $y = {x^2} + 3x - 10$?

Answer 1

Hint: In this question, we need to find the vertex and intercept of the given equation. Note that the given equation represents a parabola. Firstly, to obtain the $x$-intercept, we set the value of y equal to zero and find the point. Then, to obtain the $y$-intercept, we set the value of x equal to zero and find the point. To find the vertex of a parabola, we convert the equation of the form given by $y = a{(x - h)^2} + k$, where $(h,k)$ is the vertex of the parabola.

Complete step-by-step answer:
Given an equation of the form $y = {x^2} + 3x - 10$ …… (1)
We are asked to find the vertex and the intercepts of the above expression given by the equation (1).
Note that the equation represents the parabola which does not pass through the origin.
Now we find the intercepts on the coordinate axes, we set one of the variables as zero and find the value of the other variable.
Finding the $x$-intercept :
Taking $y = 0$ in the equation (1) we get,
$ \Rightarrow 0 = {x^2} + 3x - 10$
This can be written as,
$ \Rightarrow {x^2} + 3x - 10 = 0$
We can split the middle term as, $3x = 5x - 2x$
Hence we get,
$ \Rightarrow {x^2} + 5x - 2x - 10 = 0$
Taking out the common terms we get,
$ \Rightarrow x(x + 5) - 2(x + 5) = 0$
$ \Rightarrow (x - 2)(x + 5) = 0$
$ \Rightarrow x - 2 = 0$ or $x + 5 = 0$
Solving for x, we get,
$ \Rightarrow x = 2, $ $x = - 5$
So the x-intercepts are $x = 2,$ $x = - 5$
Finding the $y$-intercept :,
Taking $x = 0$ in the equation (1) we get,
$ \Rightarrow y = {0^2} + 3(0) - 10$
This can be written as,
$ \Rightarrow y = 0 + 0 - 10$
$ \Rightarrow y = - 10$
So the y-intercept is $y = - 10$.
The standard form of an equation of a parabola is given by $y = a{x^2} + bx + c$
Here we have $a = 1,$ $b = 3$ and $c = - 10$.
Now we make the equation of the parabola given a perfect square. We find the last term using the formula $\dfrac{{{b^2}}}{{4a}}$
Substituting the values, we get,
$ \Rightarrow \dfrac{{{3^2}}}{{4 \times 1}} = \dfrac{9}{4}$
Now we add and subtract $\dfrac{9}{4}$ in the equation (1), we get,
$ \Rightarrow y = {x^2} + 3x - 10 + \dfrac{9}{4} - \dfrac{9}{4}$
Rearranging the terms we get,
$ \Rightarrow y = {x^2} + 3x + \dfrac{9}{4} - 10 - \dfrac{9}{4}$
This can be written as,
$ \Rightarrow y = {x^2} + 2\left( {\dfrac{3}{2}} \right)x + {\left( {\dfrac{3}{2}} \right)^2} - 10 - \dfrac{9}{4}$
So simplifying this we get,
$ \Rightarrow y = {\left( {x + \dfrac{3}{2}} \right)^2} - 10 + \dfrac{9}{4}$
Taking LCM for the last terms we get,
$ \Rightarrow y = {\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{{49}}{4}$
Now we compare the above equation with the vertex form given by,
$y = a{(x - h)^2} + k$, where $(h,k)$ is the vertex of the parabola.
Here we have $a = 1$, $h = - \dfrac{3}{2}$ and $k = - \dfrac{{49}}{4}$
Hence the vertex of the parabola is $\left( { - \dfrac{3}{2}, - \dfrac{{49}}{4}} \right)$.

Note:
Students must remember that to obtain the $x$-intercept, we set the value of y equal to zero and find the point. Then, to obtain the $y$-intercept, we set the value of x equal to zero and find the point. So the simplification must be done carefully to obtain the required intercepts.
Students must know the equations representing standard form and vertex form of the parabola. Since they may get confused while solving such problems. This applies only if the given equation is a quadratic equation.
The standard form of the parabola is given by the equation, $a{x^2} + bx + c$
To find the vertex of a parabola, we use the vertex form given by, $y = a{(x - h)^2} + k$, where $(h,k)$ is the vertex of the parabola.