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# How do you find the horizontal intercept of the equation $y = - 4{x^2} - 8x + 12$ algebraically?

Last updated date: 13th Jun 2024
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Hint: We have been given to find the horizontal intercept. Horizontal intercept is the $x - axis$. Put the value of $y = 0$ in the given equation. Then, we will get the equation in the terms of $x$ - intercept and solve it by splitting the middle term.

Complete step by step solution:
From the question, we know that we have to find the horizontal intercept of the equation $y = - 4{x^2} - 8x + 12$ by the method of algebra. We know that the horizontal intercept is the intercept of the $x - axis$ whereas the vertical intercept is the $y - axis$. So, we have to find the equation which gives us the value of $x$.
Every point on the $x - axis$ has the form of $\left( {x,0} \right)$ , while every point on the graph has the form $\left( {x,f\left( x \right)} \right)$, as $f\left( x \right) = y$, so, to find the $x$ intercept we have to put the value of $y = 0$.
Therefore, putting $y = 0$ in the above equation, we get –
$\Rightarrow 0 = - 4{x^2} - 8x + 12 \\ \Rightarrow 4{x^2} + 8x - 12 = 0 \\$
Now, taking 4 as common from the above equation to make the equation more simplified –
$\Rightarrow {x^2} + 2x - 3 = 0 \cdots \left( 1 \right)$
Now, we have three parts –
The first term, ${x^2}$, has the coefficient as 1.
The middle term, $2x$, has the coefficient as 2.
The last term, $- 3$, has the coefficient as -3.
Multiply the coefficient of the first term which is 1 with the last term or constant which is -3 –
$\Rightarrow 1 \times \left( { - 3} \right) = - 3$
Now, the factors of -3 are -1 and 3 which when multiplied gives the result as -3 and sum as 2 –
$\Rightarrow - 1 + 3 = 2$
Putting these values in the equation (1), we get –
$\Rightarrow {x^2} + 3x - x - 3 = 0$
Taking like terms from the first two-term and taking common factor from the last two terms –
$\Rightarrow x\left( {x + 3} \right) - 1\left( {x + 3} \right) = 0$
Now, $\left( {x + 3} \right)$ is common in the above equation –
$\Rightarrow \left( {x + 3} \right)\left( {x - 1} \right) = 0$
When $\left( {x + 3} \right) = 0$ then, $x = - 3$
When $\left( {x - 1} \right) = 0$ then, $x = 1$

Hence, the horizontal intercepts of the equation are $\left( { - 3,0} \right)$ and $\left( {1,0} \right)$.

Note:
We can also find the horizontal intercepts by finding the discriminant and then nature of roots. Then, use the quadratic formula for the equation ${x^2} + 2x - 3 = 0$ and find the values of $x$ by using the formula –
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
where, $a$ is the coefficient of the first term, $b$ is the coefficient of the middle term and $c$ is the constant term.